A 60 N force is applied horizontally to an object that is accelerated on a frictionless horizontal surface. The mass of the object is 14 kg

Question

(a) if the object starts from rest and the force is applied for 9.0 s then how much work is done by the force?

(b) if the object starts from rest and the force is applied until the object's velocity is 6.0 m/s then how much work is done by the force?

Answer 1

(a)

Get the work from the change in kinetic energy. But first, find the final speed.

v = v₀ + at

v = v₀ + (F/m)t

v₀ = 0 (started from rest)

v = (F/m)t

Plug in the given values of F, m, and t, and get v in m/s.

W = mv²/2 - mv₀²/2 = mv²/2 (since v₀ = 0)

W = mv² /2

Plug in the values of m and v from above, and get W in joules.

(b)

W = mv²/2 - mv₀²/2 = mv²/2 (since v₀ = 0)

W = mv² / 2

In this part of the problem, plug in the given mass m and

v = 6.0 m/s

and get W in joules.