$A$ bolt of mass $0.3 \; kg$ falls from the ceiling of an elevator moving down with a 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?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Mass of the bolt,$m = 0.3 \; kg$.
Height of the elevator,$h = 3 \; m$.
Acceleration due to gravity,$g = 9.8 \; m s^{-2}$.
Since the elevator is moving with a uniform speed,its acceleration is zero. The bolt falls relative to the elevator with an initial velocity of zero.
At the time of impact,the potential energy of the bolt relative to the floor is converted into heat energy.
Heat produced = Loss of potential energy = $mgh$.
Heat produced = $0.3 \times 9.8 \times 3 = 8.82 \; J$.
If the elevator were stationary,the relative velocity of the bolt with respect to the floor would still be zero at the start of the fall. Therefore,the heat produced would remain the same,$8.82 \; J$.

Explore More

Similar Questions

Match the items in Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Work done is zero $(a)$ By gravitational force (when body moves down)
$(2)$ Work done is positive $(b)$ Against gravitational force (when body moves up)
$(3)$ Work done is negative $(c)$ By centripetal force

$A$ block of mass $5 \ kg$ is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration $2 \ m/s^2$. The work done by tension on the block during the first $3 \ s$ is $(u = 0)$. (in $J$)

Two identical balls of mass $2 \ kg$ are moving towards each other with a velocity of $5 \ m/s$. They collide and come to rest after the collision. What is the work done by the internal forces in $J$?

Explain why total linear momentum is conserved in an elastic collision,and define inelastic collision and perfectly inelastic collision.

$A$ lead bullet penetrates into a solid object and melts. Assuming that $40 \%$ of its kinetic energy is used to heat it,the initial speed of the bullet is ............ $m \, s^{-1}$.
(Given: Initial temperature of the bullet $= 127^{\circ} C$,
Melting point of the bullet $= 327^{\circ} C$,
Latent heat of fusion of lead $= 2.5 \times 10^{4} \, J \, kg^{-1}$,
Specific heat capacity of lead $= 125 \, J \, kg^{-1} K^{-1}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo