Question

A constant force acting on a body of mass $3.0kg$ changes its speed from $2.0m$ $s^{-1}$ to $3.5m$ $s^{-1}$ in $25s$. 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 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:

$v=u+at$
$a=\frac{(v-u)}{t}$

$=\frac{(3.5-2)}{25}=0.06m$$/s^2$

As per Newton's second law of motion, force is given as:

$F=ma$

$=3\times 0.06=0.18N$

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.