# Force exerted by hand

Kindly give me the explanation of this numerical which says 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 upto $$2$$ m height further , find the magnitude of the force . $$g=10\ ms^{-2}$$.

The answer in my textbook is $$20$$ N

## 2 Answers

The first thing you have to think is that it took energy to get 2m high, so it may be useful to calculate that energy. The final energy is given by:

$$mgh = 0.2 \cdot 10 \cdot 2 J = 4 J$$

that energy is coming from the kinetic energy at the start of the motion, but that shange of speed is due to the force applied. Since the change of kinetic energy is equal to the work and the work is force times distance, we have

$$\Delta E = W = F\cdot l = F \cdot 0.2$$

via conservation of energy we have

$$mhg = F\cdot l$$

$$F = 20N$$

• hahaha you are right, is funny the more you advance into the career the more silly mistakes in this kind of things. Thank you Jul 17, 2020 at 6:15

Mass of ball, m = 0.2 Kg Acceleration due to gravity, g = 10m/s² Heigh achieved by the ball, h = 2 m Length along which force is applied by hand, s = 0.2 m Force exerted by the hand, F is to be calculated

With this data, we can calculate the force exerted by the hand using energy conservation,

mgh = Fs

(0.2)(10)(2) = F(0.2)

20 N= F

Thus force exerted by the hand is 20 N