Question

8- A girl of mass 35 kg site on trolley of mas 5kg.initial velocity of trolley is im/s Come to rest after travelling 16m.find work done on trolley.come to meet after thavelling 16 m
(a) work done ou trolley
(b) work done by girl

Answer 1

Answer:

Given:

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

Answer 2

Answer:

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Explanation:

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ANSWER

mass of girl m=35kg

mass of trolley=5kg

total mass=40kg

initial velocity of trolley u=4ms

−1

final velocity of trolley v=0ms

−1

distance s=16m

we know,

v

2

−u

2

=2as

a=

2s

v

2

−u

2

=−

2×16

16

a=−

2

1

ms

−2

Force=ma=40×(−0.5)

=-20N

work is done on the trolley=20N×16m=320 J.

work was done by the girl =0 J.