A block of mass m sits at rest on a frictionless table in a train that is moving with speed v c (w.r.t.
ground) along a straight horizontal track (fig.) A person in the train pushes on the block with a net horizontal force F for a time t in the direction of the car’s motion.

(i) What is the final speed of the block according to a person in the train?
(ii) What is the final speed of the block according to a person standing on the ground outside the train?
(iii) How much did kinetic energy of the block change according to the person in the car?
(iv) How much did kinetic energy of the block change according to the person on the ground?
(v) In terms of F, m & t how far did the force displace the object according to the person in car?
(vi) According to the person on the ground?
(vii) How much work does each say the force did?
(viii) Compare the work done to the KE gain according to each person.
(ix) What can you conclude from this computation?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) a 1 = F/m, so v 1 = a 1 t = Ft/m.
(ii) Since velocities add, v = v c + v 1 = v c + Ft/m
(iii) Δ K 1 = m(v 1 ) 2 /2 = F 2 t 2 /2m
(iv) Δ K = m (v c + v 1 ) 2 /2 – mv c 2 /2
(v) s 1 is a 1 t 2 /2 = Ft 2 /2m
(vi) s 1 + v c t
(vii) W g = F [V c t +
] , W t = F [
t 2 ]
(viii) Compare W and W 1 with Δ K and Δ K 1 , they are respectively equal.
(ix) The work - energy theorem holds for moving observers.
Sol. (i) w.r.t. person in the train
v 1 = at = 
(ii) w.r.t. person on ground,
v = v c + v 1 = v c + 
(iii) According to person in the train,
Δ K 1 =
mv 1 2 = 
(iv) According to person on ground,
Δ K =
m. 
(v) S 1 =
a 1 t 2 =
.
(vi) According to person on ground,
S = v c t +
t 2 =
+ v c t.
(vii) According to person in the train
work done by F} = Fs 1
= 
According to person on ground,
Work done by F} = F.s
=
.
(viii) Comparing W g = Δ K g
and W c = Δ K c .
(ix) Work–energy theorem holds in moving frame also.
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