Question

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 m height further, find the magnitude of the force, applied by the hand. Consider g = 10 m/$s^2$

• A. 20 N
• B. 22 N
• C. 4 N
• D. 16 N

Answer 1

The correct option is B

22 N

The ball is at rest in the starting point A. The hand accelerates the ball by applying force on it from A to B. so the ball would have achieved some velocity (u) when it reaches point B.

Now ball leaves contact of hand with velocity u and travels 2m under gravity to reach C.

At this point balls velocity is 0

for the journey from A to B initial velocity of ball $u_1$ = 0

find velocity i.e., velocity at B = u =$\sqrt{40}$

s = 0.2

40 = $u_1{}^2$ + 2 $\times$ a (0.2)

40 = 2 a 0.2 a = 100 m/$s^2$

$\therefore$ F = m $\times$ a will give the net force acting on the ball

The hand pushes the ball upward while the earth pulls is down with mg.

$\Rightarrow$ F - mg = ma

F = mg + ma

= m (g + a)

= 0.2 $\times$ (10 + 100)

F = 22 N

Alternatively to find net force

Force =$\frac{△p}{△t}$ p is momentum

From A to B

$△p$ = mv - mu

= m(v-u)

$v^2=u^2+2as$

$(\sqrt{40}{)}^2=(0{)}^2+2\times a\times 0.2$

40 = 0.4a

a = 100 m/$s^2$

$\therefore t=\frac{v-u}{a}=\frac{\sqrt{40}-0}{100}=\frac{\sqrt{40}}{100}=\frac{\sqrt{10}}{50}s$

$\therefore \frac{△p}{△t}=\frac{m(v-u)}{t}=\frac{0.2(\sqrt{40}-0)}{\sqrt{40}}\times 100$

= 20N

∴ Net force = 20 N

So the force applied by the hand=20+2=22 N