#### Question

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?

#### Solution 1

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.

#### Solution 2

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.