Work, Power and EnergyMedium
Question
A man of mass m is stationary on a stationary flat car. The car can move without friction along horizontal rails. The man starts walking with velocity v relative to the car, work done by him
A man of mass m is stationary on a stationary flat car. The car can move without friction along horizontal rails. The man starts walking with velocity v relative to the car, work done by him
A man of mass m is stationary on a stationary flat car. The car can move without friction along horizontal rails. The man starts walking with velocity v relative to the car, work done by him
Options
A.is less than
mv2 if he walks along the rails
B.is equal to
mv2 if he walks normal to the rails
C.can never be less than
mv2
D.is greater than
mv2 if he walks along the rails
Solution
If the man walks along the rails, some velocity say V is imparted to car also. Let M be the mass of car. Then from conservation of linear momentum.
M.V = m(v – V)
∴ V =
∴ work done by man
=
m (v − V)2 +
mV2
=
v2 <
mv2
Hence, option (A) is correct. If the man moves normal to the rails then car will not move. Hence, work done by him in the case will be
mv2 and option (B) is also correct.
Mathod II
Man of the mass m initially at rest on the statianary car of mass M. Now man starts walking (Let parallel to the rail) then car moves with vC
Working from the graund
MVC = m(v − VC) ⇒ VC =
velocity of the car towards Left hand side

Work done by the man = (ᐃK)man + (ᐃK)car
=
m (V − VC)2 +
MV2C
=
m
+
MV2C
=
(M + m) V2
=
<
MV2 
M.V = m(v – V)
∴ V =
∴ work done by man
=
=
Hence, option (A) is correct. If the man moves normal to the rails then car will not move. Hence, work done by him in the case will be
Mathod II
Man of the mass m initially at rest on the statianary car of mass M. Now man starts walking (Let parallel to the rail) then car moves with vC
Working from the graund
MVC = m(v − VC) ⇒ VC =

Work done by the man = (ᐃK)man + (ᐃK)car
=
=
=
=
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