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.
Solution 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 = F/m = (-8.0)/(0.40) = -20 "m/s"^2`
(i) At t = -5s
Acceleration, a' = 0 and u = 10 m/s
`s = ut + 1/2 a't^2`
= 10 × (–5) = –50 m
(ii) At t = 25 s
Acceleration, a'' = –20 m/s2 and u = 10 m/s
s' = ut' + 1/2 a"t2
`= 10 xx 25 + 1/2 xx (-20) xx (25)^2`
= 250 + 6250 =- 6000 m
(iii) At t = 100 s
For `0 <= t <= 30`
a = –20 m/s2
u = 10 m/s
S1 = ut + 1/2 a"t2
`= 10 xx 30 + 1/2 xx (-20) xx (30)^2`
= 300 - 9000
= -8700 m
For 30' < t <= 100s
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:
s2 = vt + 1/2 a"t2
= –590 × 70 = –41300 m
∴Total distance, S" = s1 + s2 = -8700 - 41300 = -50000 m
Solution 2
Here m = 0.40 kg, u = `10 ms^(-1)`, F = -8 N(retarding force)
As `a = F/m = - 8/0.4 = - 20 ms^(-2)`
Also `S = ut + 1/2 at^2`
1) Position at t = - 5s
`S = 10(-5) + 1/2 xx 0 xx (-5)^2 = -50 m`
2) Position at t = 25 s
`S_1 = 10 xx 25 + 1/2 xx (-20)xx(25)^2 = -6000 m = - 6 km`
3) Postion at t = 100 s
`S_2 = 10xx30 + 1/2 xx (-20) xx (30)^2 = -8700 m`
At t = 30 s, v = u + at
`v = 10 - 20 xx 30 = -590 ms^(-1)`
Now for motion from 30 s to 100s
`S_3 = -590 xx 70 + 1/2 (0) xx (70)^2 = -41300 m`
`"Total distance" = S_2 + S_3 = -8700 - 41300 = -50000 m = - 50 km`
