1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 15 InterviewSolution

701.	A particle of mass m initially at rest. A variabl force acts on the particle `f=kx^(2)` where k is a constant and x is the displacment. Find the work done by the force f when the speed of particles is v.A. `1/2kx^(2)`B. `1/2kv^(2)`C. `1/2mv^(2)`D. `2mv^(2)`
Answer» Correct Answer - C

Discussion

702.	A force is given by `F=kx^(2)` where x is in meters and `k=10 N//m^(2)` What is the work done by this force when it acts from x=0 to x=0.1m ?
Answer» Correct Answer - `[3.33 xx 10^(-3) J]`

Discussion

703.	A block of mass m is released from a height h from the top of a smooth surface. There is an ideal spring of spring constant k at the bottom of the track. Find the maximum compression in the spring (Wedge is fixed) A. `sqrt((2mgh)/(k))`B. `sqrt((mgh)/(k))`C. `sqrt((mgh)/(2k))`D. Cannot be determined
Answer» Correct Answer - A

Discussion

704.	A particle of mass m initially moving with speed v.A force acts on the particle f=kx where x is the distance travelled by the particle and k is constant. Find the speed of the particle when the work done by the force equals W.A. `sqrt((k)/(m)+v^(2))`B. `sqrt((2W)/(m)+v^(2))`C. `sqrt((2W)/(k)+v^(2))`D. `sqrt((W)/(2m)+v^(2))`
Answer» Correct Answer - B

Discussion

705.	Consider the system shown in the figure. Coefficient of friction between the block and table is `mu=0.5.` The system is released from rest. Find the work done by friction, when the speed of block is 10 m/s `(m=1kg)` A. 100 JB. `-100J`C. 200 JD. `-200 J`
Answer» Correct Answer - C

Discussion

706.	A train is moving with a speed of `90 kmh^(-1)` passenger X inside the train displaces his 40 kg, luggage slowly an the floor through 1 m in 10 s. Coefficient of friction of the floor of the train is 0.2. Find the work done by this passenger X and the luggage as seen by (i) a follow passenger Y (ii) a person on the ground [Take ,g=10 `ms^(-1)`].
Answer» Given, speed of the train, `v=90kmh^(-1) =(90xx1000)/(60xx60)=25 ms^(-1)` (i) Displacement of the luggage with respect to the train,S=1 m. As luggage is displaced slowlt, the force applied on it must be same as frictional force on it by the floor `f=mu" mg"=(0.2)xx40xx10=80 N` Work done by the passenger X as seen by fellow passenger Y is `W =F_(s)=(80)(1)=80 J` (ii) The luggage is dispalced for 10 s. Therefore distance moved by train with respect to ground during this interval is `s_(0)=25xx10=250 m` Therefore work done by passenger X on the luggage as seen by a person on the ground is `W_(G)=F(s+s_(0))=(80)(1+250)=(251)(80)=20.08 kJ`

Discussion

707.	Consider the system shown in the figure. Coefficient of friction between the block and table is `mu=0.5.` The system is released from rest. Find the work done by friction, when the speed of block is 10 m/s `(m=1kg)` A. `-10J`B. `-20J`C. `-30J`D. `-50J`
Answer» Correct Answer - D

Discussion

708.	Consider the spring block system shown in the figure. Initially block m is in equilibrium. Then block m is displaced b y a small distance h. Find the maximum speed of the block. ltbgt A. `hsqrt((k)/(m))`B. `2hsqrt((k)/(h))`C. `hsqrt((m)/(k))`D. `2hsqrt((m)/(k))`
Answer» Correct Answer - A

Discussion

709.	Consider a block connected to a light spring of spring constant `100Nm^(-1)`. Now the block is displaced by applying a constant force F which gives zero resultant force when spring is stretched through 10 cm. Evaluate (i) Work done by the spring force when the block attains equilibrium. (ii) Net work done on the block when it attains maximum speed.
Answer» (i) Work done by the spring force, `W=(1)/(2) kx^(2)` `k=100 Nm^(-1) " and " x=10 cm=0.1 m` `:.W_(s)=(1)/(2)xx100(0.1)^(2)=-0.5 J` Negative sign indicates that the work done by the spring force is negative. (ii) When the block attains equilibrium, its speed is maximum. `:.` Work done on the block by external force F `W=F.x=(10)((10)/(100))=1 J` Thus, net work done on the block `W_(N)=W_(S)+W=-0.5+1 =0.5 J`

Discussion

710.	A toy rocket of mass 0.1 kg hass a small fuel of mass 0.02 kg, which it burns out in 3s. Starting from rest on a horizontal smooth track, it gets a speed of `20ms^(-1)` after the fuel is burnt out. What is the approcimate thrust of the rocket ? What is the energy content per unit mass of the fuel ? (Ignore the small mass variation of teh rocket during fuel burning).
Answer» Here,`M=0.1kg, m=0.02kg, t=3s,` `u=0,upsilon=20m//s, Thrust=?, en ergy//mass=?` Thrust of rocket, `F=Ma=(M(upsilon-u))/(t)=(0.1(20-0))/(3)=(2)/(3)N` KE gained by rocket `=(1)/(2)Mupsilon^(2)=(1)/(2)xx0.1(20)^(2)` `=20J` Energy `//` mass of fuel`=(20)/(0.02)=1000Jkg^(-1)`

Discussion

711.	In which of the following cases can the work done increase the potential energy?A. (a) Both conservative and non-conservative forcesB. (b) Conservative force onlyC. (c) Non-conservative force onlyD. (d) Neither conservative nor non-conservative forces.
Answer» Correct Answer - B In case of non-conservative forces, the work done is dissipated as heat, sound, etc., i.e., it does not increase the potential energy. But in case of conservative forces, work done is responsible for increasing or decreasing the potential energy.

Discussion

712.	Consider the two cases: a. A says that the potential energy of a book placed on a shelf is `10J`. b. B says that the potential energy of the same book is `-10J`. Which one of them is wrong? Justify.
Answer» Yes. One of them is wrong. Change in potential energy is calculated by considering the initial and the final positions of the body. In this case, the initial position would be the ground and the final position would be the elevated position. Therefore, B is wrong. Also change in potential energy is independent of reference level taken.

Discussion

713.	A metre stick weighing 600 g, is displaced through an angle of `60^@` in vertical plane as shown. The change in its potential energy is (`g = 10 m s^(-2)`) A. 1.5 JB. 15 JC. 30 JD. 45 J
Answer» Correct Answer - A `DeltaU=(MgL)/2 (1-cos theta)=(600xx10xx1)/(1000xx2)xx1/2` `DeltaU=6/4=1.5 J`

Discussion

714.	Under the action of a force, a 1 kg body moves, such that its position x as function of time t is given by `x=(t^(3))/(2).` where x is in meter and t is in second. The work done by the force in fiest 3 second isA. `143 J`B. `2430 J`C. `(729)/(8)J`D. `24.3 J`
Answer» Correct Answer - C

Discussion

715.	Work done by a frictional force isA. Is always zeroB. is always negativeC. Is always positiveD. May be positive, negative or zero
Answer» Correct Answer - D

Discussion

716.	Assertion : Work done by the frictional force is negaive Reason : This is because frictional force acts along the dirction of motionA. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - C Frictional force always opposes relative motion. Therefore, work done by the frictional force is negative. Assertion is true, reason is false.

Discussion

717.	A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to `(-K//r^(2))`, where k is a constant. The total energy of the particle is -A. `-(K)/(r )`B. `(1.5k)/(r )`C. `-(K)/(2r)`D. `-(3k)/(2r)`
Answer» Correct Answer - C

Discussion

718.	Assertion : Work done by the centripetal force in moving a body along a circle is always zero. Reason : Because displacement of the body is along the force.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - C Work done by centripetal force is zero, because displacement of body is perpendicular to the direction of force, `W=F s cos 90^(@)= zero`

Discussion

719.	Assertion , No work is done if the displacement is zero Reason: Work done by the force is defined to be the product of component of the force in the the direction of the displacement and the magnitude of displacement.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true not but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false.
Answer» Correct Answer - A Work done `W=Fd cos theta` if displacement is zero, i.e. d=0, W = 0

Discussion

720.	Which among the followinig, is a form of energyA. LightB. PressureC. MomentumD. Power
Answer» Correct Answer - A

Discussion

721.	A body is heated by giving amount of heat energy? Will its mass increase or decrease or remain constant? If it increase or decrease then by how much?
Answer» The variations in mass depends on the nature of body. For most solids, there is no variations in mass of heating. However, if the solid is volatile then on heating some loss of mass occurs due to evaporation.

Discussion

722.	A football kicked by a player leaves the ground and after travelling in air reached the ground and comes to stop at some other position on the ground. Identify the energy transformation in this process. Is energy of the ball conserved?
Answer» Kinetic energy imparted to the ball is converted into potential energy at the highest point and small part of it remains kinetic. Then it is converted into kinetics. Then it is converted into kinetic energy when it hits the ground. This energy is used up in overcoming the friction against air and the ground. Due to dissipation of energy in overcoming friction against air and ground, the kinetic energy of the ball is not conserved.

Discussion

723.	A particle is acted by `x` force `F = Kx` where `K` is `a( + Ve)` constant its potential mwrgy at `x = 0` is zero . Which curve correctly represent the variation of putential energy of the block with repect to `x`A. B. C. D.
Answer» Correct Answer - B (b) WE know that `delta U = -W` for conservative forces `deltaU = - int_(0)^(x) Fdx ` or `deltaU = - int_(0)^(x) k xdx` `rArr U_(x) - U_(0) = (kx^(2))/(2)` Given ` U_(0) = 0 U_(x) = -(kx^(2))/(2)`

Discussion

724.	A mass m starting from A reaches B of a frictionless track. On reaching B, it pushes the track with a force equal to x times its weight, then the applicable relation is A. (a) `h=((x+5))/(2)r`B. (b) `h=x/2r`C. (c) `h=r`D. (d) `h=((x+1)/(2))r`
Answer» Correct Answer - A KE of blocks at `B=PE` at `A-PE` at B `1/2mv^2=mgh-mg2r=mg(h-2r)` `v^2=2g(h-2r)` (i) Also, `(mv^2)/(r)=xmg+mg` or `v^2=(x+1)rg` (ii) Equating Eqs (i) and (ii), we get `2g(h-2r)=(x+1)gr` or `2gh=(x+1)gr+4gr=(x+5)gr` `h=((x+5)/(2))r`

Discussion

725.	A body of mass `50 kg` is projected vertically upward with velocity of `100 m//s`. After `5 s` this body breaks into `20 kg` and `30 kg`. If the `20 kg` piece travels upwards with `150 m//s`, then the velocity of other block will beA. 15m/sec downwardsB. 15m/sec upwardsC. 51m/sec downwardsD. 51m/sec upwards
Answer» Correct Answer - A

Discussion

726.	A body is moving with a velocity v , breaks up into two equal parts. One of the part retraces back with velocity v . Then the velocity of the other part isA. v in forward directionB. 3v in forward directionC. v in backward directionD. 3v in backward direction.
Answer» Correct Answer - B

Discussion

727.	What is the minimum energy released in the annhilation of an electron positron pair?
Answer» The rest mass of an electron or positron is `m_(0)=9.1xx10^(-31)kg.` In the annhilation of an electron positron pair, the rest mass of the two particles is converted into energy. `:.` Minimum energy released, `E=2m_(0)c^(2)=2(9.1xx10^(-31))(3xx10^(8))^(2)jou l e`. ` E=(2xx9.1xx9xx10^(-15))/(1.602xx10^(-13))MeV=1.02MeV`

Discussion

728.	A projectile is fired with some velocity making certain angle with the horizontal. Which of the following graphs is the best representation for the kinetic energy of a projectile (KE) versus its horizontal displacement(x)?A. (a) B. (b) C. (c) D. (d)
Answer» Correct Answer - D Velocity of a projectile at any instant of time(t) is `V^2=v_x^2+v_y^2=(u cos theta)^2+(u sin theta-g(x)/(u cos theta))^2` `:. KE=1/2m u^2-mgx tan theta+(mg^2x^2)/(u^2cos^2theta)` The given equation represents the equation of a parabola.

Discussion

729.	Referring the graphs, which of the following is/are correc? A. (a) The particle has stable equilibrium at point 3 and b.B. (b) The particle is in neutral equilibrium at point b and 2.C. (c) No power is delivered by the force on the particle at point 1, 3, and b.D. (d) The particle has least kinetic energy at position 1.
Answer» Correct Answer - A::C::D Option (a) is correct because at 3, force is opposite to displacement. At (b), `u` is minimum. Option (b) is incorrect at 2, there is net force and hence no equilibrium. Option (c) is correct because at `1,3` and (b) force is zero. At (b), `(du)/(dx)=-F=0` Option (d) is correct. Positive work is done after position 1, hence KE will increase. So KE is least at 1.

Discussion

730.	An adult weighing 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, the energy utilised by him in jogging isA. `9xx10^6` JB. `9xx10^5` JC. `6xx10^6` JD. `6xx10^5` J
Answer» Correct Answer - B Here, mg = 600 N, h = 0.25 m Length of each step = 1 m `therefore` No of steps in 6 km is `N=(6xx10^3m)/(1 m) =6000` `therefore` Energy utilised in jogging is `E=Nmgh = 6000 xx 600 N xx 0.25 m = 9 xx 10^(5)J`

Discussion

731.	An adult weighing 600N raises the centre of gravity of his body by 0.25m while taking each step of 1m lenth in jogging. If he jogs for 6km, calculate the energy utilised by him in jogging assuming that there is no energy loss dur to friction of ground and air. Assuming that the body of the adult is capable of converting `10%` of energy intake in the form of food, calculate the energy equivalent fo food that would be required to compensate energy utilised for jogging.
Answer» Given, weight of the adult(w)=mg=600N Height of each step=h=0.25 Length of each step=1m Total distance travelled=6km=6000m Total number of steps=`(6000)/(1)=6000` Total energy utlised in jogging=`nxxmgh` `=6000xx699xxC.25=9xx10^(5)J` Since, 10% of intake energy is utilised in jogging Total intake energy=`10xx9xx10^(5)J=9xx10^(6)J` Total intake energy `=10xx9xx10^(5)J=9xx10^(6)J`

Discussion

732.	The energy required to accelerate a car from 10 m/s to 20 m / s is how many times the energy required to accelerate the car from rest to 10 m /sA. Equal to the acceleration of the particleB. 4 timesC. 2 timesD. 3 times
Answer» Correct Answer - D

Discussion

733.	The work done by a force `vec(F)=(-6x^(3)hat(i))` N in displacing a particle from x = 4m to x = - 2m isA. `-240` JB. 360 JC. 420 JD. Will depend upon the path
Answer» Correct Answer - B

Discussion

734.	Give reason for the following: Pardeep says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with him? Why?
Answer» Yes, I agree with Pardeep. According to newton’s law of motion for a body: Net force = mass × acceleration So, Acceleration = \(\frac{Net\,force}{Mass}\) Every object on the Earth possesses some mass. So, if the resultant or net force due to several forces acting on the object is zero then the acceleration of the body will be zero by the above-mentioned formula.

734.

Give reason for the following: Pardeep says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with him? Why?

Answer»

Yes, I agree with Pardeep. According to newton’s law of motion for a body:

Net force = mass × acceleration

So, Acceleration = \(\frac{Net\,force}{Mass}\)

Every object on the Earth possesses some mass. So, if the resultant or net force due to several forces acting on the object is zero then the acceleration of the body will be zero by the above-mentioned formula.

Discussion

735.	Two blocks 1 and 2 start from same point A on a smooth slide at the same time. The track from A to B to C is common for the two blocks. At C the track divides into two parts. Block 1 takes the route C–D–E and gets airborne after E. Block 2 moves along CFGH. Point E is vertically above G and the stretch GH is horizontal. Block 1 lands at point H. (a) Where is the other block at the time block 1 lands at H? Has it already crossed H or yet to reach there? (b) Which block will reach at H with higher speed ?
Answer» Correct Answer - (a) Block 2 has already corssed H. (b) Both reach H with same speed.

Discussion

736.	A bob of mass m, suspended by a string of length `l_1` is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length `l_2`, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio `(l_1)/(l_2)` is
Answer» Correct Answer - `(5)` The initial speed of 1st bob (suspended by a string of length `l_1`) is `sqrt(5gl_1)`. The speed of this bob at height point will be `sqrt(gl_1)` When this bob collides with the other bob there speeds will be interchanged. `sqrt(gl_1)=sqrt(5gl_1)impliesl_1/l_2=5`

Discussion

737.	Two bodies of mass `1 kg and 2 kg` have equal momentum. The ratio of their kinetic energies is:A. `1:1`B. `2:1`C. `1:3`D. `3:1`
Answer» Correct Answer - D We know that `KE = (P^(2))/(2m)` Hence, for same momentum, `KE prop (1)/(m)`

Discussion

738.	Assertion: Universe as a whole may be viewed an isolted system. Reason: Total energy of an isolated system remain constant or stretched.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false.
Answer» Universe as a whole may be viewed as an isolated system. The total energy of the universe is constant. If one part of the universe loses energy, another part must gain an equal amount of energy.

Discussion

739.	A particle of mass 0.5 kg travels in a straight line with velocity v = a x3/2 where a = 5 m-1/2s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Answer» Given m = 0.5 kg. V = ax^3/2, a = 5m^-1/2 s^-1 initial velocity at x = 0, v₁= a × 0 = 0 Final velocity at x = 2n is, V₂ = a (2)^3/3 = 5 × 2^3/2 work done = change in KE \(\frac {1}{2}\)m (v₂² – v₁²) = \(\frac {1}{2}\)× 0.5 × 25 × 2³ = 50 J .

739.

A particle of mass 0.5 kg travels in a straight line with velocity v = a x3/2 where a = 5 m-1/2s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Answer»

Given m = 0.5 kg.

V = ax^3/2, a = 5m^-1/2 s^-1

initial velocity at x = 0, v₁= a × 0 = 0

Final velocity at x = 2n is,

V₂ = a (2)^3/3 = 5 × 2^3/2

work done = change in KE

\(\frac {1}{2}\)m (v₂² – v₁²) = \(\frac {1}{2}\)× 0.5 × 25 × 2³

= 50 J .

Discussion

740.	A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1 . What is the speed of the trolley after the entire sand bag is empty?
Answer» The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sandbag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, the speed of the trolley will remain 27 km/h.

Discussion

741.	A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Answer» Mass of the bolt, m = 0.3 kg Speed of the elevator = 7 m/s Height, h = 3 m Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. Heat produced = Loss of potential energy = mgh = 0.3 x 9.8 x 3 = 8.82 J The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

741.

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Answer»

Mass of the bolt, m = 0.3 kg
Speed of the elevator = 7 m/s

Height, h = 3 m
Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.
Heat produced = Loss of potential energy
= mgh = 0.3 x 9.8 x 3
= 8.82 J

The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

Discussion

742.	Which of the following potential energy curves in figure., cannot possibley describly describe the elastic collision of two billiard balls ? Here r is distance between centres of the balls.
Answer» The potential energy of a system of two masses varies inveresly as the distance ® between them, i.e., `V(r)prop(1)/(r)`. When the twor billiard balls touch eachother,P.E. becomes zero i.e. at `r=R+R=2R,V(r )=0`. Out of the given curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two biliard balls.

Discussion

743.	Which of the following potential energy curves in figure., cannot possibley describly describe the elastic collision of two billiard balls ? Here r is distance between centres of the balls. A. B. C. D.
Answer» Correct Answer - C The potential energy of a system of two masses varies inversely as the distance (r) between them. i.e. `V(r)prop 1/r` When the two billiard balls touch each other potential energy becomes zero, i.e at r=R+R=2R,V(r)=0 Out of the given graphs, only curve (c ) satisfies these conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.

Discussion

744.	Which of the following potential energy curves in figure., cannot possibley describly describe the elastic collision of two billiard balls ? Here r is distance between centres of the balls.
Answer» The potential energy of a system of two masses varies inversely as the distance(r) between 1 them, i.e., `V(r)alpha 1//r`. When the two billard touch each other, P.E. Becomes zero i.e.,at r=R+R=2R, V(r) =0. Out of the given graphs, curve (v) only satisfies these two condition. Therefore, all other curves cannot possibly describe the eleastic collision of two billiard balls.

Discussion

745.	State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.
Answer» (i) If the displacement of the body is in the direction of force, then work done is positive. Hence, W= F x S For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction. (ii) If the displacement of the body is in the direction opposite to the force, then work done is negative. Hence, W = − F × S For example: When a body moves on a surface, the force of friction between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.

745.

State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.

Answer»

(i) If the displacement of the body is in the direction of force, then work done is positive.

Hence, W= F x S

For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.

(ii) If the displacement of the body is in the direction opposite to the force, then work done is negative.

Hence, W = − F × S

For example: When a body moves on a surface, the force of friction between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.

Discussion

746.	Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature ? Why?
Answer» No. W.D. is zero only in case of a conservative force.

Discussion

747.	A toy is acted upon by a force. State two conditions under which the work done could be zero.
Answer» We know that W = FScosθ. clearly, work done by the force will be zero if. 1. S = 0 i.e., no displacement takes place 2. If θ = 90°, then cos 90° = 0, and W = 0;.e., when the force acts perpendicular to the direction of motion.

747.

A toy is acted upon by a force. State two conditions under which the work done could be zero.

Answer»

We know that W = FScosθ. clearly, work done by the force will be zero if.

1. S = 0 i.e., no displacement takes place

2. If θ = 90°, then cos 90° = 0, and W = 0;.e., when the force acts perpendicular to the direction of motion.

Discussion

748.	State the two conditions under which a force does no work ?
Answer» (i) Displacement is zero or it is perpendicular to force. (ii) Conservative force moves a body over a closed path.

748.

State the two conditions under which a force does no work ?

Answer»

(i) Displacement is zero or it is perpendicular to force.

(ii) Conservative force moves a body over a closed path.

Discussion

749.

Give the Various conditions for vertical motion.

Answer»

Various conditions for vertical motion :

Velocity at lowest point	Condition
u_A > √5gl	Tension in the string will not be zero at any of the point and body will continue the circular motion.
u_A = √5gl	Tension at highest point C will be zero and body will just complete the circle.
√2gl < u_A < √5gl	Particle will not follow circular motion. Tension in string become zero somewhere between points B and C whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory.
u_A = √2gl	Both velocity and tension in the string becomes zero between A and B and particle will oscillate along semi-circular path.
u_A < √2gl	Velocity of particle becomes zero between A and B but tension will not be zero and the particle will oscillate about the point A.

Discussion

750.	The sign of work done by a force on a body is important to understand. State carefully if the following quantities are poisitve or negative : (a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) Work done by gravitational force in the above case. (c ) Work done by friction on a body sliding down an iniclined plane. (d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity. (e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer» Work done `W=vecF. vecs=Fs cos theta` where `theta` is smaller angle between force `vec F` and displacement `vec s`. (a) To lift the bucket, force equal to weight of bucket has to be applied vertically upwards and the bucket moves along the same direction i.e., `theta=0^(@)` therefore `W=F s cos 0^(@)=F s.` It si positive. (b) The bucket moves in a direction opposite to the gravitational force which acts vertically downwards. `:. theta=180^(@), W=F s cos 180^(@)=-Fs.` It is negative. (c) Friction always opposes the relative motion `:. theta =180^(@), W=F s cos 180^(@)=-Fs`. It is negative. (d) As the body moves along the direction of applied force `theta=0^(@), :. W=F s cos 0^(@)=Fs.`. It is positive. (e) The direction of resistive force is opposite to the direction of motion of the bob, i.e., `theta=180^(@).`. Hence, work done, in this case, is negative.

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