Question

A body of mass 0.40 kg moving initially with a constant speed of 10 m s-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

Given Data

• Mass of body: \( m = 0.40 \,\text{kg} \)
• Initial speed (towards north, taken as positive \(x\)-direction): \( u = 10 \,\text{m s}^{-1} \)
• Force applied towards south (negative \(x\)-direction): \( F = -8.0 \,\text{N} \)
• Duration of force: \( 0 \le t \le 30 \,\text{s} \)
• At \( t = 0 \), position \( x = 0 \)

Acceleration (for \( 0 \le t \le 30 \,\text{s} \))

\( a = \dfrac{F}{m} = \dfrac{-8.0}{0.40} = -20 \,\text{m s}^{-2} \)

1. Position at \( t = -5 \,\text{s} \)

For \( t < 0 \), the force has not yet been applied, so the body moves with uniform velocity \( u = 10 \,\text{m s}^{-1} \) (no acceleration).

\( x(t) = x_0 + u t = 0 + 10 \times (-5) = -50 \,\text{m} \)

Position at \( t = -5 \,\text{s} \): \( x = -50 \,\text{m} \) (50 m south of the origin).

2. Position at \( t = 25 \,\text{s} \) (within the force-acting interval)

For \( 0 \le t \le 30 \,\text{s} \), use uniformly accelerated motion:

\( x(t) = ut + \dfrac{1}{2} a t^{2} \) \( x(25) = 10 \times 25 + \dfrac{1}{2} (-20) \times 25^{2} \) \( = 250 - 10 \times 625 = 250 - 6250 = -6000 \,\text{m} \)

Position at \( t = 25 \,\text{s} \): \( x = -6000 \,\text{m} = -6.0 \,\text{km} \) (6 km south of origin).

3. Position at \( t = 100 \,\text{s} \)

(i) Motion from \( t = 0 \) to \( t = 30 \,\text{s} \)

Displacement in first 30 s:

\( x_1 = ut + \dfrac{1}{2} a t^{2} = 10 \times 30 + \dfrac{1}{2} (-20) \times 30^{2} \) \( = 300 - 10 \times 900 = 300 - 9000 = -8700 \,\text{m} \)

Velocity at \( t = 30 \,\text{s} \):

\( v = u + a t = 10 + (-20) \times 30 = 10 - 600 = -590 \,\text{m s}^{-1} \)

(ii) Motion from \( t = 30 \) to \( 100 \,\text{s} \)

After 30 s, the force is no longer acting, so the body moves with constant velocity \( v = -590 \,\text{m s}^{-1} \) for the next \( 70 \,\text{s} \).

\( x_2 = v \Delta t = (-590) \times 70 = -41300 \,\text{m} \)

(iii) Total displacement from \( t = 0 \) to \( t = 100 \,\text{s} \)

\( x_{\text{total}} = x_1 + x_2 = -8700 + (-41300) = -50000 \,\text{m} = -50 \,\text{km} \)

Position at \( t = 100 \,\text{s} \): \( x = -50 \,\text{km} \) (50 km south of the origin).