A girl having mass of 35 KG sits on a trolley of mass 5 KG. The trolley is given an initial velocity of 4 m/s by applying a force. The trolley comes to rest after travelling a distance of 16 m.
(i) How much work is done on the trolley?
(ii) How much work is done by the girl?
3 answers
in each case, the object traveling has initial kinetic energy, and work is done by each on friction that equals this KE.
Mass of girl = 35 kg
Mass of trolley = 5 kg
Total mass, m = (35 + 5) = 40 kg
u = 4 m/s
v = 0
s = 16 m
By equation,
v 2 = u 2 + 2as
0 = (4)2 + 2a (16)
32a = –16
a = –0.5 m/s2
Force exerted on trolley,
F = ma
= 40 × 0.5 = 20 N
Work done on trolley W = FS = 20 N x 16 m
= 320 J
Work done by the girl, W = FS
= mass of girl × retardation × S
= 35 × 0.5 × 16
= 280 J
Mass of trolley = 5 kg
Total mass, m = (35 + 5) = 40 kg
u = 4 m/s
v = 0
s = 16 m
By equation,
v 2 = u 2 + 2as
0 = (4)2 + 2a (16)
32a = –16
a = –0.5 m/s2
Force exerted on trolley,
F = ma
= 40 × 0.5 = 20 N
Work done on trolley W = FS = 20 N x 16 m
= 320 J
Work done by the girl, W = FS
= mass of girl × retardation × S
= 35 × 0.5 × 16
= 280 J
work done by girl is zero,because she is sitting on the trolley and force is perpendicular to displacement.
but in book it is given 280J.
How is it?
but in book it is given 280J.
How is it?
3 answers