The speed of a particle is doubled, its kinetic energy:A. remains same

1.	The speed of a particle is doubled, its kinetic energy:A. remains same B. becomes double C. becomes four times D. becomes half
Answer» Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula: KE = \(\frac{1}{2}\) mv² so kinetic energy is directly proportional to the mass of object and to the square of speed of the body. If the speed of particle is doubled then kinetic energy becomes four time. KE₁ =\(\frac{1}{2}\) mv₁² KE₂ = \(\frac{1}{2}\) mv₂² where, v₂ = 2v₁ Substituting the value of v₂ in the formula, KE₂= \(\frac{1}{2}\) m (2v₁)² = \(\frac{1}{2}\)(4 mv₁² ) KE₂= 4 KE₁ So kinetic energy will become 4 times if the speed is doubled.

The speed of a particle is doubled, its kinetic energy:A. remains same B. becomes double C. becomes four times D. becomes half

Answer»

Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula:

KE = \(\frac{1}{2}\) mv²

so kinetic energy is directly proportional to the mass of object and to the square of speed of the body.

If the speed of particle is doubled then kinetic energy becomes four time.

KE₁ =\(\frac{1}{2}\) mv₁²

KE₂ = \(\frac{1}{2}\) mv₂² where, v₂ = 2v₁

Substituting the value of v₂ in the formula,

KE₂= \(\frac{1}{2}\) m (2v₁)²

= \(\frac{1}{2}\)(4 mv₁² )

KE₂= 4 KE₁

So kinetic energy will become 4 times if the speed is doubled.

Discussion

No Comment Found

Related InterviewSolutions

A 3-kg steel ball strikes a wall with a speed of `10.0ms^(-1)` at an angle of `60.0^(@)` with the surfaces of the wall. The ball bounces off with the same speed and same angle. If the ball was in contact with the wall for 0.2s, find the average force exerted by the wall on the ball. A. `300 N`B. zeroC. `150 sqrt(3) N`D. `150N`
The distance x moved by a body of mass 0.5 kg under the action of a force varies with time t as`x(m)=3t^(2)+4t+5` Here, t is expressed in second. What is the work done by the force in first 2 seconds?
The power (P) of an engine lifting a mass of 100 kg upto a height of 10 m in 1 min isA. P=163.3 WB. P=9800 WC. P=10000 WD. P=5000 W
A weight lifter lifts a weight off the ground and holds is up, thenA. work is done in lifting as well as holding the weight.B. no work is done in both lifting and holding the weight.C. work is done in lifting the weight but no work is required to done in holding it up.D. no work is done in lifting the weight but work is required to be done in holding it up.
Calculate the worked done in lifting a 300N weight to a height of 10m with an accelertaion `0.5ms^(2)`. Take g=`10ms^(-2)`.
A perosn is holding a bucket by applying a force of 20N. He moves a horizontal distance of 15m first and then climbs up a distance of 12m. What is the total work done by him?
What is tidal energy ? How is it harnessed ?
A weight lifter does……………in lifting the weight………………..the ground but ……………in………………..the weight up.
While catching a cricket ball of mass `200g` moving with a velocity of `20ms^(-1)`, the player draws his hands backwards through `20cm`. Find the work done in catching the ball and the average force exerted by the ball on the hand.
Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.