Question

14. 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 16m.
a) How much work is done on the trolley?
b) How much work is done by the girl?

Answer 1

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:320 joule and 280 joule.

Explanation:from work energy theorem work done by the system = change in kinetic energy.

(a) total mass of the system =m1 +m2

(35+ 5)kg = 40 kg.

Initial velocity u = 4m/s

Final velocity v=0m/s

Hence work done on the trolly =(m1 +m2 )×[u^2-v^2]/2

W= 40×[4^2-0^2]/2 joule

W=40×16/2 joule

W=40×8 joule

W=320 joule.

(b) work done by the girl = 35×[u^2-v^2]/2

W=35×[4^2-0^2]/2

W=35×16/2

W=35×8joule

W=280 joule