Question

A body of mass 0.40 kg moving initially with a constant speed of 10 ms^-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.

Answer 1

Here m = 0.40 kg, u = 10 ms^-1

F = -8 N (retarding force)

Therefore, a = F/m = -8/0.4 = -20 ms^-2

Form S = ut + 1/2 at²

(i) Position, at t = -5 s

S_-5 = 10 x (-5) + 1/2 x 0 x (-5)² = -50 m

(ii) Position, at t = 25 s

S₂₅ = 10 x 25 + 1/2 x (-20) x (25)²

= -6000 m = -6 km

(iii) Position, at t = 100 s

Let us first consider motion upto 30 seconds

S₃₀ = 10 x 30 + 1/2 x (-20) x (30)²

and = -8700 m

At t = 30

v₃₀ = u + at = 10 + (-20) x (30) = -590 ms^-1

Force balance 70 s

S_30-100 = -590 x 70 + 1/2 x 0 x (70)² = -41300 m

Total distance = S₃₀ + S_30-100

= -41300 - 8700 = -50,000 m = -50 km