Question

A constant force acting on a body of mass 3.0 $\text {kg}$ changes its speed from 2.0 $\text m \, \text s^{-1}$ to 3.5 $\text m \, \text s^{-1}$ in 25 $\text s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?

Answer 1

Mass of the body, m = 3 $\text {kg}$
Initial speed of the body, u = 2 $\text m / \text s$
Final speed of the body, v = 3.5 $\text m / \text s$
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 = $\frac{\text v-\text u}{\text t}$

a = $\frac{3.5-2}{25}$

a = $\frac{1.5}{25}$

a = 0.06 $\text m / \text s^2$
As per Newton’s second law of motion, force is given as:
F = ma
F = 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.