Question

$A.$ A constant retarding force of 50 N applied to a body of mass 20 Kg moving initially with a speed of 15 m $s^{-1}$. How long does the body take to stop?
$B.$ A constant force acting on a body of mass 3.0 Kg change 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 remain same, the magnitude of the force will be

Answer 1

$A.$ Given that,

Force $F=50N$

Mass $m=20kg$

Initial velocity $u=15m/s$

Now, from newton’s second law

$F=ma$

$50=20\times a$

$a=2.5m/s^2$

Now from equation of motion

$v=u+at$

$0=15+(-2.5)t$

$t=\frac{15}{2.5}$

$t=6s$

Hence, the body will be stop in 6 sec.

$B.$ Given that,

Mass of the body,$m=3kg$

Initial speed of the body $u=2m/s$

Final speed of the body $v=3.5m/s$

Time, $t=25sec$

Now, for acceleration

Use the equation of motion

$v=u+at$

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

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

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

$a=0.06m/s^2$

Now, for the force

Use the newton’s second law

$F=ma$

$F=3\times 0.06$

$F=0.18N$

The magnitude of the force is $0.18N$.