Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Work Power and Energy question

2021 · 1 Sep · Shift 2 · Q50
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Work Power and Energy
  5. /2021 · 1 Sep · Shift 2 · Q50

Work Power and Energy question

2021 · 1 Sep · Shift 2 · Q50

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8 gh\sqrt {gh}gh​. The value of workdone by the air-friction is :
  1. A
    −-− 0.68 mgh
  2. B
    mgh
  3. C
    1.64 mgh
  4. D
    0.64 mgh
View written solutionFree

Correct answer: A

  1. Use the work-energy theorem

When the body is dropped from height hhh:

  • Initial kinetic energy: Ki=0K_i = 0Ki​=0
  • Initial potential energy: Ui=mghU_i = mghUi​=mgh
  • Final potential energy at ground: Uf=0U_f = 0Uf​=0
  • Final speed is given as v=0.8ghv = 0.8\sqrt{gh}v=0.8gh​

So final kinetic energy is

Kf=12mv2=12m(0.8gh)2K_f = \frac{1}{2}mv^2 = \frac{1}{2}m\left(0.8\sqrt{gh}\right)^2Kf​=21​mv2=21​m(0.8gh​)2
  1. Compute the final kinetic energy

Since

(0.8)2=0.64(0.8)^2 = 0.64(0.8)2=0.64

we get

Kf=12m(0.64gh)=0.32mghK_f = \frac{1}{2}m(0.64gh) = 0.32mghKf​=21​m(0.64gh)=0.32mgh
  1. Relate this to work done by air friction

Loss in gravitational potential energy is mghmghmgh.

If there were no air resistance, final kinetic energy would be mghmghmgh. But actually, the body has only

0.32mgh0.32mgh0.32mgh

of kinetic energy.

Hence, the remaining energy has been dissipated by air friction.

Using work-energy theorem:

Wg+Wair=ΔKW_g + W_{\text{air}} = \Delta KWg​+Wair​=ΔK

Here,

  • Work done by gravity: Wg=mghW_g = mghWg​=mgh
  • Change in kinetic energy: ΔK=0.32mgh\Delta K = 0.32mghΔK=0.32mgh

Thus,

mgh+Wair=0.32mghmgh + W_{\text{air}} = 0.32mghmgh+Wair​=0.32mgh

So,

Wair=0.32mgh−mgh=−0.68mghW_{\text{air}} = 0.32mgh - mgh = -0.68mghWair​=0.32mgh−mgh=−0.68mgh
  1. Match with the options
Wair=−0.68mghW_{\text{air}} = -0.68mghWair​=−0.68mgh

So the correct option is A.

PreviousNext

More from Work Power and Energy

  • An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of 72 kmh − 1 when the brakes are applied to bring it to rest. In the process of the system being…2021 · Numerical
  • A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?2021 · MCQ
  • As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s2) The value… Includes diagram2021 · Numerical
  • A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :-2021 · MCQ
  • A ball of mass 4 kg, moving with a velocity of 10 ms − 1, collides with a spring of length 8 m and force constant 100 Nm − 1. The length of the compressed spring is x m. The value of x, to the nearest integer, is ​…2021 · Numerical
  • In a spring gun having spring constant 100 N/m a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball… Includes diagram2021 · Numerical
  • A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to :2021 · MCQ
  • A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (take g = 9.8 ms − 2)2021 · MCQ