Question

Two constant forces are applied to a 5 kg block that is initially at rest on a frictionless and horizontal surface.

12 N 2 N

What is the displacement of the object 2 s after the forces are applied?

What is work done by the 2 N force during the first 2 s after the forces are applied?

What is change in the kinetic energy of the block during the first 2 s after the forces are applied?

Answer 1

(a) Resultant force on the block, F = 12 - 2 = 10 N

So, acceleration of the block, a = F / m = 10 / 5 = 2 m/s^2

So, displacement of the block after t = 2 s -

s = u*t + (1/2)*a*t^2 = 0 + 0.5*2*2^2 = 4 m.

(b) Work done by 2 N force, W = F*s = 2 * (-4) = -8.0 J (negative sign shows that the work is done opposite to the direction of the force)

(c) Velocity of the block after 2 s -

v = u + a*t = 0 + 2*2 = 4 m/s

So, change in the kinetic energy of the block during first 2 s -

KE = (1/2)*m(v2)^2 - (1/2)*m(v1)^2 = 0.5*5*4^2 - 0 = 40 J.