1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 14 InterviewSolution

651.	The potential energy of a spring, when stretched through a distance x is 50J. What would be the work done in stretchin it further through the same distance?
Answer» Here, `x_(1=)x, E_(1)=(1)/(2)kx^(2)=50J, W=?x_(2)=x+x=2x` As work done `=` increase in P.E. of the spring `:.W=E_(2)-E_(1)=(1)/(2)kx_(2)^(2)-(1)/(2)kx_(1)^(2)=(1)/(2)k[(2x)^(2)-x^(2)]W=3xx(1)/(2)kx^(2)=3xx50=150J`

Discussion

652.	In the arrangement shown in figure `m_(A)=4.kg` and `m_(B)=1.0kg`. The system is released from rest and block B is found to have a speed `0.3m//s` after it has descended through a distance of `1.m` find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take `g=10 m//s^(2)`). .
Answer» From constant relations, we can see that `v_(A)=2v_(B)` Therefore, `" " v_(A)=2(0.3)=0.6 ms^(-1)` as `" " v_(B)=0.3 ms^(-1) " "`(given) Applying `" " W_(nc)=DeltaU+DeltaK` we get `" " -mu m_(A)gS_(A)=-m_(B)gS_(B)+(1)/(2)m_(A)v_(A)^(2)+(1)/(2)m_(B)v_(B)^(2)` Here `" " S_(A)=2S_(B)=2m` as `" " S_(B)=1 m " " `(given) `:.-mu(4.0)(10)(2)=-(1)(10)(1)+(1)/(2)(4)(0.6)^(2)+(1)/(2)(1)(0.3)^(2)` or `" " -80 mu= -10+0.72+0.045` or `" " 80 mu=9.235` `" or " mu=0.115`

Discussion

653.	The length of a steel wire increases by `05cm` , when it is loaded with a weight of `50kg`. Calculate force constant of the wire and workdown in stretching the wire. Take `g=10ms^(-1)`.
Answer» Here, `x=05cm=5xx10^(-3)m, F=50kg=50N` Force constant, `K=(F)/(x)=(50)/(5xx10^(-3)=10^(4)N//m` Work done in stretching the wire `W=(1)/(2)Kx^(2)=(1)/(2)xx10^(4)(5xx10^(-3))^(2)=(0.25)/(2)=0.125J`

Discussion

654.	In an amusement park passengers riding in a Air racer are revolving around a tall steel tower. At top speed all of its planes fly at `60^(@)` bank, and about 50 m from the tower. In this position the support chains make an angle of `60^(@)` with the vertical Calculate the speed of the planes.
Answer» Correct Answer - B::D

Discussion

655.	A rectangular block is mobing along a frictionless path, when it encounters the circular loop as shown. The block passes points `1,2,3,4,` before ret runing to the horizontal tracks. At point 3 A. Its mechanical energy is minimumB. The foeces on it area balancedC. It experiences a net upward forceD. Its speed is minimum
Answer» Correct Answer - D

Discussion

656.	Consider the following statements A and B. Identify the correct choice in the given answers A. In a one dimensional perfectly elastic collision between two moving bodies of equal masses the bodies merely exchange their velocities after collision. B. If a lighter body at rest suffers perfectly elastic collision with a very heavy body moving with a certain velocity, then after collision both travel with same velocity.A. A and B are correctB. Both A and B are wrongC. A is correct, B is wrongD. A is wrong, B is correct
Answer» Correct Answer - C For A: Using the laws of conservation of linear momentum and energy, the velocities of two bodies after perfectly elastic collision are given by `v_1=((m_1+m_2)u_1)/(m_1+m_2)+(2m_2u_2)/(m_1+m_2)`...(i) `v_2=(2m_1u_1)/(m_1+m_2)+((m_2+m_1)u_2)/(m_1+m_2)`...(ii) where `m_1` and `m_2` are the masses of two bodies and `u_1` and `u_2` are the velocities of two bodies before collision. If `m_1= m_2`, then From (i) and (ii), we get `v_1=u_2 and v_2=u_1` i.e., in perfectly elastic collision of two moving bodies ofequal masses, the bodies merely exchange their velocities after collision. Thus, statement A is correct. For B: In Eqs. (i) and (i), If `m_2lt lt m_1` and `u_2= 0`, then `v_1=u_1 and v_2 = 2u_1` Thus, statement B is wrong.

Discussion

657.	What average power is genereted by a `90.0 kg` mountain climbs who climbs a summit of height `600 m` in `90.0 min`?A. `100W`B. `50 W`C. `25 W`D. `200W`
Answer» Correct Answer - A We assume the climber has negligible speed at , both the beginning and the end of the climb .Then `K_(f) = K_(p)` and the work done by the muscles is `W_("ne") = 0 + (U_(f)- U_(i)) = mg(y_(f) - y_(i))` `= (90.0 kg) (10.0m//s^(2))(600m)` ` = 5.40 xx 10^(5)J` The evarege the power delivered is `P= (W_(nc))/(Delta t)= (5.40 xx 10^(5)J)/((90 min)(60s//1min)) = 100W`

Discussion

658.	If `W_(1) W_(2)` and `W_(3)` represent the work done in moving a particle from `A` to `B` along three different paths `1.2` and`3` respectively (asshown ) in the gravitational fieled of a point mass m, find the correct relation between ` `W_(1) W_(2)` and `W_(3)` A. `W_1gtW_2gtW_3`B. `W_1=W_2=W_3`C. `W_1ltW_2ltW_3`D. `W_2gtW_1gtW_3`
Answer» Correct Answer - B Work done by a conservative force is independent of path followed by particle.

Discussion

659.	If `W_(1) W_(2)` and `W_(3)` represent the work done in moving a particle from `A` to `B` along three different paths `1.2` and`3` respectively (asshown ) in the gravitational fieled of a point mass m, find the correct relation between ` `W_(1) W_(2)` and `W_(3)` A. `W_(1) gt W_(2) gt W_(3)`B. `W_(1)=W_(2)=W_(3)`C. `W_(1) lt W_(2) lt W_(3)`D. `W_(2) gt W_(1) gt W_(3)`
Answer» Correct Answer - B

Discussion

660.	A particle of mass m1 is moving with a velocity `v_(1)` and another particle of mass `m_(2)` is moving with a velocity v2. Both of them have the same momentum but their different kinetic energies are E1 and E2 respectively. If `m_(1) gt m_(2)` thenA. `E_(1) lt E_(2)`B. `(E_(1))/(E_(2))=(m_(1))/(m_(2))`C. `E_(1) gt E_(2)`D. `E_(1)=E_(2)`
Answer» Correct Answer - A

Discussion

661.	A particle of mass m1 is moving with a velocity `v_(1)` and another particle of mass `m_(2)` is moving with a velocity v2. Both of them have the same momentum but their different kinetic energies are E1 and E2 respectively. If `m_(1) gt m_(2)` thenA. `E_(1)ltE_(2)`B. `(E_(1))/(E_(2))=(m_(1))/(m_(2))`C. `E_(1)gtE_(2)`D. `E_(1)=E_(2)`
Answer» Correct Answer - a

Discussion

662.	A ball of mass`2 kg` and another of mass `4 kg` are dropped together from a `60` feet tall building . After a fall of `30` feet each towards earth , their respective kinetic energies will be the ratio ofA. `sqrt(2):1`B. `1:4`C. `1:2`D. `1:sqrt(2)`
Answer» Correct Answer - C

Discussion

663.	A ball of mass`2 kg` and another of mass `4 kg` are dropped together from a `60` feet tall building . After a fall of `30` feet each towards earth , their respective kinetic energies will be the ratio ofA. `sqrt2:1`B. `1:4`C. `1:2`D. `1:sqrt2`
Answer» Correct Answer - c

Discussion

664.	which ball can reach D?A. 1B. 2C. 1 and 2D. Cannot be predicted
Answer» Correct Answer - B Ball 1 loses energy by friction. It cannot cross point C.

Discussion

665.	A small roller coaster starts at point A with a speed u on a curved track as shown in the figure The friction between the roller coaster and the track is negligible and it always remains in contact with the track. The speed of roller coaster at point D on the track will beA. `(u^2 + gh)^(1//2)`B. `(u^2 + 2gh)^(1//2)`C. `(u^2 + 4gh)^(1//2)`D. u
Answer» Correct Answer - C Let the speed of the roller coaster at point D be v. Applying the law of conservation of energy between points A and D, we get `1/2m u^2 +mg(2h)=1/2mv^2+0` or `u^2 +4gh=v^2` or `v=sqrt(u^2+4gh)`

Discussion

666.	The force constant of a wire is k and that of another wire is . 2k When both the wires are stretched through same distance, then the work doneA. `W_(2)=2W_(1)^(2)`B. `W_(2)=2W_(1)`C. `W_(2)=W_(1)`D. `W_(2)=0.5W_(1)`
Answer» Correct Answer - B

Discussion

667.	Force constant of two wires `A` and `B` of the same material are `K` and `2K` respectively. If the two wires are stretched equally, then the ratio of work done in stretching `((W_(A))/(W_(B)))` isA. `(1)/(3)`B. `(1)/(3)`C. `(1)/(2)`D. `(1)/(4)`
Answer» Correct Answer - B (b) We know that the work done ina stretched wire `W=(1)/(2)kx^(2)` Given,`" " k_(A)=k and k_(B)=2k` So that`" " W_(A)=(1)/(2)kx^(2) and W_(B)=(1)/(2)(2k)x^(2)=kx^(2)` Hence, the ratio of work done in stretch wire `(W_(A))/(W_(B))=((1//2)kx^(2))/(kx^(2))rArr(W_(A))/(W_(B))=(1)/(2)`

Discussion

668.	When does a force do work?
Answer» Work is said to be done only when the force applied on a body makes the body moves he., there is displacement of the body.

668.

When does a force do work?

Answer»

Work is said to be done only when the force applied on a body makes the body moves he., there is displacement of the body.

Discussion

669.	The power of a heart which pumps `5 xx 10^(3) cc` of blood per minute at a pressure of `120 mm` of meteury `(g = 10 ms^(-2)` and density of `Hg = 13.6 xx 10^(3) km^(3))` isA. `1.50`B. `1.70`C. `2.35`D. `3.0`
Answer» Pressure `= 150 mm Hg` Pumping rate `= (dV)/(dt) = (5 xx 10^(-3))/(60) m^(3)//s` Power of heart `P. (dV)/(dt) = rho gh xx (dV)/(dt)` `= (13.6 xx 10^(3) kg//m^(3)) (10) xx (0.15) xx (5 xx 10^(-3))/(60)` `= (13.6 xx 5 xx 0.15)/(6) = 1.70` watt.

Discussion

670.	What is the work done by the moon when it revolves around the earth?
Answer» No work is done as moon revolves around the earth in Circular Path and angle between force acting towards the center of circle and tangent at any point in circular path is 90°.

Discussion

671.	(a) Define work. (b) What are the conditions for doing work (c) State the mathematical expression for work.
Answer» (a) “When force is applied on the body and body moves (covers , some distance) in the direction of force, work is said to be done.” Or “Work is said to be done, when a force or its component causes a displacement in its own direction.” (b) (1) Force should be applied. (2) Displacement of body should be there. (C) Work = Force x displacement W = F x S

671.

(a) Define work. (b) What are the conditions for doing work (c) State the mathematical expression for work.

Answer»

(a) “When force is applied on the body and body moves (covers , some distance) in the direction of force, work is said to be done.”

Or

“Work is said to be done, when a force or its component causes a displacement in its own direction.”

(b)

(1) Force should be applied.

(2) Displacement of body should be there.

(C) Work = Force x displacement

W = F x S

Discussion

672.	An ice cream has a marked value of 700 kcal. How may kilowatt-hour of energy will it deliver to the body as it is degested.A. 0.81 kWhB. 0.90 kWhC. 1.11 kWhD. 0.71 kWh
Answer» Correct Answer - A

Discussion

673.	Define 1 Kilowatt hour.
Answer» 1 KWh : 1 KWh is the amount of energy consumed when an electric appliance having a power rating of 1 Kilowatt is used for 1 hour.

Discussion

674.	A bulb of 60 Watt is used for 6 hrs. daily. How many units (KWh) of electrical energy are consumed ?
Answer» Power of bulb = 60 W = 60/1000 KW = 0.06 KW t = 6 hours Energy = Power × Time taken = 0.06 × 6 h = 0.36 KWh = 0.36 units

674.

A bulb of 60 Watt is used for 6 hrs. daily. How many units (KWh) of electrical energy are consumed ?

Answer»

Power of bulb = 60 W

= 60/1000 KW

= 0.06 KW

t = 6 hours

Energy = Power × Time taken

= 0.06 × 6 h

= 0.36 KWh

= 0.36 units

Discussion

675.	The conversion of part of the energy into an undesirable form is called.
Answer» The conversion of part of the energy into an undesirable form is called Dissipation of energy.

Discussion

676.	Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .A. `1.6 J`B. `16 J`C. `160 J`D. `1600 J`
Answer» Correct Answer - B `v = (dx)/(dt) = (d)/(dt) ((t^(3))/(3)) = t^(2)` Where `t = 0` When `t = 2`, then `t = 4 m//s` Work down in first two second = change in KE `W = (1)/(2)m [(4)^(2) - (0)^(2)] = (1)/(2) 2 xx 16 = 16`

Discussion

677.	The displacement `x` of particle moving in one dimension, under the action of a constant force is related to the time `t` by the equation ` t = sqrt(x) +3` where `x is in meters and t in seconds` . Find (i) The displacement of the particle when its velocity is zero , and (ii) The work done by the force in the first ` 6 seconds`.A. `18 m`B. zeroC. `9 m//2`D. `36 m`
Answer» Correct Answer - B Here `t = sqrtx+ 3 or x = (t - 3)^(2) = t^(2) - 6t + 9` `v = (dx)/(dt) = 2t + 6` at `t = 0, v = 2 xx 0 - 6 = - 6` at `t = 6 s, v = 2 xx 6 - 6 = +6` Initial and final `KE` ar esame hence no work is done.

Discussion

678.	A 10 kg mass moves x-axis. Its acceleration as function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x = 0 to x = 8 cm ? A. `8xx10^(-2)`joulesB. `16xx10^(-2)`joulesC. `4xx10^(-4)` jouleD. `1.6xx10^(-3)` joules
Answer» Correct Answer - A

Discussion

679.	Can whole of the kinetic energy be lost in a completely inelastic collision?
Answer» Yes, if the total momentum of particles before collision is zero.

Discussion

680.	How are the velocities of the particles of same mass related when they collide in a linear elastic collision?
Answer» After collision, the velocities of the particles are interchanged.

Discussion

681.	Define newton, in SI unit of force. State its relationship with CGS unit of force.
Answer» 1. One newton is that force which when acting on a body of m ass 1kg, produces an acceleration of 1ms^-2 in it. 2. S.I. unit of force is newton (N) and C.G.S. unit is dyne. IN = 10⁵ dyne.11.

681.

Define newton, in SI unit of force. State its relationship with CGS unit of force.

Answer»

1. One newton is that force which when acting on a body of m ass 1kg, produces an acceleration of 1ms^-2 in it.

2. S.I. unit of force is newton (N) and C.G.S. unit is dyne. IN = 10⁵ dyne.11.

Discussion

682.	Body `A `of mass `m` and `B` of mass `3m` move towards each other with velocities `V and 2 V` respectivally from the positions as shown , along a smooth horizontal circule track of redius `r`. After the first elastic collision , they will collide again after the time: A. `(2 pi r)/(V)`B. `(pi r)/(2V)`C. `(pi r)/(V)`D. `(2 pi r)/(3V)`
Answer» Correct Answer - B `e = 1 = (Velocity of separation)/(2 V - (- V))` Velocity of separation `= 3 V` Required `= (2 pi r)/(3 V)`

Discussion

683.	A weightlifter lift a weight off the ground and bolds is up, thenA. work is done in lifting as well as holding the weightB. No work is done in both lifting and holding the weightC. work is done in lifting the weight but no work is required by donein holding it upD. no work is done in lifting the weight but work is required to be done in holding it up
Answer» Correct Answer - C When a weight lifter lifts a weight by beight h(say).then work done by the lifting force F : `W_(1) = Fs cos 0^(@) = + Fs` but work done in holding it up is zero because the displacement is zero

Discussion

684.	Which of the following is//are conservative force (s) ?A. (a) `vecF=2r^3vecr`B. (b) `vecF=-5/rhatr`C. (c) `vecF=(3(xhati+yhatj))/((x^2+y^2)^(3//2))`D. (d) `vecF=(3(yhati+xhatj))/((x^2+y^2)^(3//2)`
Answer» Correct Answer - A::C Since, `W=intvecFdvecr` Clearly for forces (a) and (b), the integration does not require For c. `W_(C)=int (3(xhati+yhatj))/((x^2+y^2)^(3//2))(dxhati+dyhatj)=3int(xdx+ydy)/((x^2+y^2)^(3//2))` Taking `x^2+y^2=t` , we get `2xdx+2ydy=dt` `xdx=ydy=(dt)/(2)` `W_c=3int(dt//2)/(t^3//2)=3/2int(dt)/(t^(3//2))` which will be independent of path after solving. Hence, a., b., and c., are conservative forces. But d. requires some more information on path to solve it. Hence, it is non-conservative.

Discussion

685.	Choose the correct statement(s) from the following.A. (a) Force acting on a particle for equal time intervals can produce the same change in momentum but different change in kinetic energy.B. (b) Force acting on a particle for equal displacemenets can produce same change in kinetic energy but different change in momentum.C. (c) Force acting on a particle for equal time intervals can produce different change in momentum but same change in kinetic energy.D. (d) Force acting on a particle for equal displacements can produce different change in kinetic energy but same change in momentum.
Answer» Correct Answer - A::B The change in momentum for each time interval `t_0` is `Deltap=Ft_0` which is constant. But the change in kinetic energy in each time interval `t_0` is different. First `t_0` interval: `DeltaK_1=(Deltap^2)/(2m)=(F^2t_0^2)/(2m)` Secon `t_0` interval: `DeltaK_2=((2Deltap)^2-(Deltap)^2)/(2m)=(3F^2t_0^2)/(2m)` The change in kinetic energy is `DeltaK=Fx_0=const ant`. First `x_0` displacement: `Deltap_1=sqrt(2mDeltaK_1)=sqrt(2m(Fx_0))` Second `x_0` displacement: `Deltap_2=sqrt(2m(2Fx_0))-sqrt(2m(Fx_0))=0.414sqrt(2m(Fx_0))`

Discussion

686.	A heavy particle of mass m is in motion on a smooth surface of hemisphere of radius R and there is no friction. At the initial instant the particle is at the topmost point A and has an initial velocity `v_(0)` At what point will the particle leave the surface of the hemisphere ? Also determine the value of `v_(0)` for which the particle will leave the sphere at the initial instant.
Answer» Correct Answer - A::B::C

Discussion

687.	A body of mass 0.8 kg has intial velocity `(3hati-4hatj) ms^(-1)` and final velocity `(-6hatj+2hatk) ms^(-1)`. Find change in kinetic energy of the body?
Answer» Change in kinetic energy `DeltaKE=(1)/(2)mv_(f)^(2)-(1)/(2)mv_(i)^(2)` where, `v_(f)=sqrt(6^(2)+4^(2))=sqrt(40)` and `v_(i)=sqrt(3^(2)+4^(2))=sqrt(25)` `(1)/(2)xx0.8[sqrt(40))^(2)-(sqrt(25)^(2))]` `=0.4[40-25]=0.4(15)=6 J`

Discussion

688.	A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to (a) v(b) v2(c) v3(d) v4
Answer» Correct answer is (c) V⁴

Discussion

689.	In which directions do these forces act?
Answer» Forces acting on both directions, upward direction and downward direction.

Discussion

690.	If the potential energy of the particle is \(a- \frac{β}{2}x^2\) , then force experienced by the particle is(a) \(β = \frac{β}{2}x^2\)(b) F = βx(c) F = -βx(d) F = -\(\frac{β}{2}x^2\)
Answer» Correct answer is (c) F = -βx

Discussion

691.	A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.(a) Who pushed a greater distance here? (b) What about the force applied (c) Who did the greater work? (d) Which is the factor influencing work here?
Answer» (a) Second boy (b) Force is same (c) Second boy done greater work (d) The Factor influencing work is displacement The factor affecting work done are force (F) and displacement (S) Equation for calculating work done Work W = Fs Unit of work is Joule (J), Kilojoule (KJ) 1KJ = 1000 J If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

691.

A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.(a) Who pushed a greater distance here? (b) What about the force applied (c) Who did the greater work? (d) Which is the factor influencing work here?

Answer»

(a) Second boy

(b) Force is same

(c) Second boy done greater work

(d) The Factor influencing work is displacement

The factor affecting work done are force (F) and displacement (S)

Equation for calculating work done Work W = Fs

Unit of work is Joule (J), Kilojoule (KJ) 1KJ = 1000 J

If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

Discussion

692.	A body of mass m kg is placed on a table. What are the forces experienced by this body?
Answer» Weight of the body applies downwards and the table applies an equal force upwards.

Discussion

693.	A body of mass 4kg initially at rest is subjected to a force 16 N. What is the kinetic energy acquired by the body at the end of 10s?
Answer» Given m = 4kg, u = 0 F = 16 N, t = 10 s F = ma 16 = 4a, a = 4ms^-2 We know that V = u + at = 0 + 4 × 10 V = 40ms^-1.

693.

A body of mass 4kg initially at rest is subjected to a force 16 N. What is the kinetic energy acquired by the body at the end of 10s?

Answer»

Given m = 4kg, u = 0

F = 16 N, t = 10 s F = ma

16 = 4a, a = 4ms^-2

We know that V = u + at = 0 + 4 × 10

V = 40ms^-1.

Discussion

694.	What energy transformation takes place just before the flower pot reaches the ground?
Answer» Potential energy is converted completely into kinetic energy

Discussion

695.	These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five reponses (a) If both Assertion and Reason arecorrect and Reason is the correct explanation of Asserrtion. (b) If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. (c ) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. Assertion If work done by conservative force is negative, then potential energy associated with that force should increase. Reason This is according to the relation `DeltaU=-W` Here, `DeltaU` is change in potential energy and W is work done by conservative force.
Answer» Correct Answer - A (a) For conservative forces, `DeltaU=-DeltaW` where `Delta`U=change in PE `Delta`W=work done by conservative force.

Discussion

696.	A particle of mass m, attached to the end of string of length l is released from the initial position A as shown in the figures. The particle moves in a vertical circular path about O. When it is vetically below O, the string makes contact with nail N placed directly below O at distance h and rotat es around it. If the particle just complete the vertical circle about N,then A. `h=(3l)/(5)`B. `h=(2l)/(5)`C. `h=(l)/(5)`D. `h=(4l)/(5)`
Answer» Correct Answer - D

Discussion

697.	A system consists of t wo identical masses A and B of mass m each connected to ends of a massless spring of force constant k as shown in figure. The minimum force F applied vertically downward on A, such that on its release, B will leave the floor.A. mgB. 2mgC. 3mgD. 4mg
Answer» Correct Answer - B

Discussion

698.	Consider the body of mass 0.2 kg and a smooth incline of hight 3.2 m and length 10 m. Select the correct alternativeA. Minimum work required to lift the block from the ground and put it at the top is 6.4 JB. Work required to slide the body up the incline (slowly) is 6.4 JC. Maximum speed of the body slipping down from rest on the plane, on reaching the ground is 8 m/sD. Work required to slide the body down the plane is more than 6.4 J
Answer» Correct Answer - A::B::C

Discussion

699.	A body of mass 2 kg is dropped from rest from a height 20 m from the surface of Earth. The body hiys the ground with velocity 10 m/s, then work done `(g=10m//s^(2))`A. On the body is 100 jB. By the gravity is 400 JC. By the dissipative force is 300 JD. By the disspipative force is `-300J`
Answer» Correct Answer - A::B::D

Discussion

700.	An object of mass m is released from rest from the top of a smooth inclined plane of height h. Its speed at the bottom of the plane is proportional toA. `m^0`B. `m`C. `m^2`D. `m^(-1)`
Answer» Correct Answer - A Let v be the speed of the object at the bottom of the plane. According to work-energy theorem `W = DeltaK = K_f- K_i` `mgh=1/2mv^2-1/2m u^2 " " (therefore u=0)` `mgh=1/2 mv^2 "or" v=sqrt(2gh)` From this expression, it is clear that v is independent of the mass of an object.

Discussion

Explore topic-wise InterviewSolutions in .

The potential energy of a spring, when stretched through a distance x is 50J. What would be the work done in stretchin it further through the same distance?

The length of a steel wire increases by `0*5cm` , when it is loaded with a weight of `5*0kg`. Calculate force constant of the wire and workdown in stretching the wire. Take `g=10ms^(-1)`.

In an amusement park passengers riding in a Air racer are revolving around a tall steel tower. At top speed all of its planes fly at `60^(@)` bank, and about 50 m from the tower. In this position the support chains make an angle of `60^(@)` with the vertical Calculate the speed of the planes.

What average power is genereted by a `90.0 kg` mountain climbs who climbs a summit of height `600 m` in `90.0 min`?A. `100W`B. `50 W`C. `25 W`D. `200W`

A ball of mass`2 kg` and another of mass `4 kg` are dropped together from a `60` feet tall building . After a fall of `30` feet each towards earth , their respective kinetic energies will be the ratio ofA. `sqrt(2):1`B. `1:4`C. `1:2`D. `1:sqrt(2)`

A ball of mass`2 kg` and another of mass `4 kg` are dropped together from a `60` feet tall building . After a fall of `30` feet each towards earth , their respective kinetic energies will be the ratio ofA. `sqrt2:1`B. `1:4`C. `1:2`D. `1:sqrt2`

which ball can reach D?A. 1B. 2C. 1 and 2D. Cannot be predicted

The force constant of a wire is k and that of another wire is . 2k When both the wires are stretched through same distance, then the work doneA. `W_(2)=2W_(1)^(2)`B. `W_(2)=2W_(1)`C. `W_(2)=W_(1)`D. `W_(2)=0.5W_(1)`

Force constant of two wires `A` and `B` of the same material are `K` and `2K` respectively. If the two wires are stretched equally, then the ratio of work done in stretching `((W_(A))/(W_(B)))` isA. `(1)/(3)`B. `(1)/(3)`C. `(1)/(2)`D. `(1)/(4)`

When does a force do work?

The power of a heart which pumps `5 xx 10^(3) cc` of blood per minute at a pressure of `120 mm` of meteury `(g = 10 ms^(-2)` and density of `Hg = 13.6 xx 10^(3) km^(3))` isA. `1.50`B. `1.70`C. `2.35`D. `3.0`

What is the work done by the moon when it revolves around the earth?

(a) Define work. (b) What are the conditions for doing work (c) State the mathematical expression for work.

An ice cream has a marked value of 700 kcal. How may kilowatt-hour of energy will it deliver to the body as it is degested.A. 0.81 kWhB. 0.90 kWhC. 1.11 kWhD. 0.71 kWh

Define 1 Kilowatt hour.

A bulb of 60 Watt is used for 6 hrs. daily. How many units (KWh) of electrical energy are consumed ?

The conversion of part of the energy into an undesirable form is called.

Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .A. `1.6 J`B. `16 J`C. `160 J`D. `1600 J`

A 10 kg mass moves x-axis. Its acceleration as function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x = 0 to x = 8 cm ? A. `8xx10^(-2)`joulesB. `16xx10^(-2)`joulesC. `4xx10^(-4)` jouleD. `1.6xx10^(-3)` joules

Can whole of the kinetic energy be lost in a completely inelastic collision?

How are the velocities of the particles of same mass related when they collide in a linear elastic collision?

Define newton, in SI unit of force. State its relationship with CGS unit of force.

Which of the following is//are conservative force (s) ?A. (a) `vecF=2r^3vecr`B. (b) `vecF=-5/rhatr`C. (c) `vecF=(3(xhati+yhatj))/((x^2+y^2)^(3//2))`D. (d) `vecF=(3(yhati+xhatj))/((x^2+y^2)^(3//2)`

A body of mass 0.8 kg has intial velocity `(3hati-4hatj) ms^(-1)` and final velocity `(-6hatj+2hatk) ms^(-1)`. Find change in kinetic energy of the body?

A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to (a) v(b) v2(c) v3(d) v4

In which directions do these forces act?

If the potential energy of the particle is \(a- \frac{β}{2}x^2\) , then force experienced by the particle is(a) \(β = \frac{β}{2}x^2\)(b) F = βx(c) F = -βx(d) F = -\(\frac{β}{2}x^2\)

A body of mass m kg is placed on a table. What are the forces experienced by this body?

A body of mass 4kg initially at rest is subjected to a force 16 N. What is the kinetic energy acquired by the body at the end of 10s?

What energy transformation takes place just before the flower pot reaches the ground?

A system consists of t wo identical masses A and B of mass m each connected to ends of a massless spring of force constant k as shown in figure. The minimum force F applied vertically downward on A, such that on its release, B will leave the floor.A. mgB. 2mgC. 3mgD. 4mg

A body of mass 2 kg is dropped from rest from a height 20 m from the surface of Earth. The body hiys the ground with velocity 10 m/s, then work done `(g=10m//s^(2))`A. On the body is 100 jB. By the gravity is 400 JC. By the dissipative force is 300 JD. By the disspipative force is `-300J`

An object of mass m is released from rest from the top of a smooth inclined plane of height h. Its speed at the bottom of the plane is proportional toA. `m^0`B. `m`C. `m^2`D. `m^(-1)`

The length of a steel wire increases by `05cm` , when it is loaded with a weight of `50kg`. Calculate force constant of the wire and workdown in stretching the wire. Take `g=10ms^(-1)`.