Question

A girl with having a 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?

Answer 1

Effective mass of girl + trolley system = mass of girl + mass of trolley = 35 kg + 5 kg = 40 kg

Here, u = 4 10m s⁻¹, v – 0, s = 16 m

Using v² – u²= 2as, we get

a = `("v"^2 - "u"^2)/(2"s")`

= `(0 - 4^2)/(2 xx 16)`

= `(-1)/2 "m s"^-2`

∴ Force, F = ma = `40 xx ((-1)/2)"N"`

= −20N

(a) Work done on the trolley = – Work done by the trolley
= − Fs
= – (–20 N) × (16 m)
= 320 J

(b) As the girl keeps sitting on the trolley, there is no displacement in her position with respect to the trolley, so no work is done by the girl.