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

Mass of the body, m = 0.40 kg 
Initial speed of the body, u = 10 m/s due north
Force acting on the body, F = – 8.0 N
Acceleration produced in the body, a =\(\frac{F}{m}\) = \(\frac{-8.0}{0.40}\) = -20 \(m/s^2\)
At t = –5 s
Acceleration, \(a\) = 0 and \(u\) = 10 \(m/s\)

s = ut + a't²

s = 10 × (–5) = –50 m
At t = 25 s 
Acceleration, \(a''\)= – 20 \(m/s^2\) and \(u\) = 10 \(m/s\)

\(s\)₁ = \(ut\) + \(\frac{1}{2}\)a''t²

= 10x25 + \(\frac{1}{2}\) x (-20) x (25)²
= 250 +6250 = -6000 m
At t = 100 s 
For 0 \(\leq\) t \(\leq\) 30 s
a = –20 \(m/s^2\)
u = 10 \(m/s\)

\(s_1\) = \(ut+\frac{1}{2}at^2\)
= 10 x 30 + \(\frac{1}{2}\) x (-20) x (30)²
= 300 - 9000
= - 8700 m
For 30'< t \(\leq\) 100 s
As per the first equation of motion, for t = 30 s, final velocity is given as:
v = u + at 
= 10 + (–20) × 30 = –590 \(m/s\)
Velocity of the body after 30 \(s\) = –590 \(m/s\)
For motion between 30 s to 100 s, i.e., in 70 \(s\):

\(s_2\) = \(vt+\frac{1}{2}\)a''t²

= –590 × 70 = – 41300 \(m\)
∴ Total distance, \(s\)‘’ = \(s_1+s_2\) = - 8700 - 41300 = - 50000 \(m\)