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– 1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?

(a)

Initial velocity of the trolley,u = 4 m/s Final velocity of the trolleyv = 0 Mass of the trolleym = 5 kg

Distance covered by the trolley before coming to rest,s = 16 m From Equation 2 as =v2-u2,

a = v2-u22 S

=0-(4)22x16

=0.5 m/s2

Force (frictional) acting on the trolley = ma

= 40 (- 0.5)

= - 20 N

Work done on the trolley = Fs = (20 N) (16 m)

= 320 J

(b) Since the girl does not move w.r.t. the trolley (as she is sitting on it),work done by the girl = 0.