A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s–1 to 3.5 m s–1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
0.18 N; in the direction of motion of the body
Mass of the body, m = 3 kg
Initial speed of the body, u = 2 m/s
Final speed of the body, v = 3.5 m/s
Time, t = 25 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
v = u + at
`:. a = (v -u)/t`
= `(3.5 - 2)/25 = 1.5/25 = 0.06 "m/s"^2`
As per Newton’s second law of motion, force is given as:
F = ma
= 3 × 0.06 = 0.18 N
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.
Here m = 3.0 kg, u = 2.0 `"ms"^(-1)`
v = 3.5 `ms^(-1)`, t = 25 s
As F = ma
or `F = m((v-u)/t)` [`:. a = (v-u)/t`]
=> F = `(3.0(3.5 - 2.0))/25 0.18 N`
The force is along the direction of motion.