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.

To solve this problem, we need to use the principle of work and energy. The work done on the ball by the hand is equal to the change in the ball's kinetic and potential energy.

1. First, we calculate the total height the ball has moved, which is the distance moved by the hand (0.2 m) plus the additional height the ball goes up (2 m). So, the total height (h) is 0.2 m + 2 m = 2.2 m.

2. Next, we calculate the work done on the ball. The work done is equal to the change in potential energy of the ball, which is mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and h is the height. So, the work done (W) is 0.2 kg * 9.8 m/s^2 * 2.2 m = 4.3 Joules.

3. The work done on the ball is also equal to the force applied by the hand times the distance the hand moved. So, we can set up the equation F*d = W, where F is the force, d is the distance moved by the hand, and W is the work done.

4. Solving for F, we get F = W/d = 4.3 Joules / 0.2 m = 21.5 N.

So, the magnitude of the force applied by the hand is 21.5 N.