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 ball goes up to 2 m height further, find the magnitude of the force. Consider g = 10 m/s^2………[?]

Options:
(a) 16 N
(b) 20 N
(c) 22 N
(d) 4 N

Answer 1

`22 N`

Explanation:

Let the ball start moving with velocity `u` and it reaches up to a maximum height `H_max`

From `H_max = u^2 / (2g)`

`u = sqrt(2gH_max) = sqrt(2 × 10 ms^-2 × 2 m) = 2sqrt(10) ms^-1`

This velocity is supplied to the ball by the hand and initially the hand was at rest, it acquires this velocity in distance of 0.2 m

`∴ a = u^2 / (2S) = (2sqrt(10) ms^-1)^2 / (2 × 0.2 m) = 100 ms^-2`

So upward force on the ball is
`F = m(g + a) = 0.2 kg (10 ms^-2 + 100 ms^-2) = 22 N`

Answer 2

See below.

Explanation:

The work/energy input to the system is

`W=E=(F-m g)delta`

This energy is completely transformed into potential energy so

`(F-mg)delta = m g h`

with

`F = m g(1+h/delta)`

here

`delta = 0.2` [m]
`m = 0.2` [K]
`h = 2` [m]

then

`F = 0.2*10(1+2/0.2) = 22` [N]