Question

4. A melon, 1.50 m above the ground, is tossed straight upward with 25 J of kinetic energy.

(a) If air resistance is negligible what is the kinetic energy of the melon when it returns to its initial level?

(b) If the melon is 2.00 kg, what is the speed when the melon leaves your hand in order to have a K.E. of 25 J?

(c) What is the total kinetic energy when the melon hits the ground?

(d) What is the speed when the melon hits the ground?

(e) Right before the melon breaks into pieces, a maximum deformation of 0.01 m is formed after it hits the ground. What is the average impact force on the melon?

Answer 1

(a) Since, air resistance is neglected and acceleration due to gravity is constant thruout the motion of the melon. then,velocity of melon in returning situation will be same but in opposite direction (downward). Thus, kinetic energy will also be same is it was initially.

Thus, kinetic energy of melon when it return to its initial level = K.E = 25J

(c) By conservation of mechanical energy,

(K.E + P.E) _{at the height 1.50m} =  (K.E+ P.E) _{at the ground ( means height = 0m)}

25J + mgh = K.E + 0 ( P.E at ground = 0 becuase height is zero)

then, K.E _{at the ground} = 25J + 2kg (9.8 m/s^2) ( 1.50) = 54.4 J Ans.

(d) again we apply the formula for kinetic energy,

(e) by eqution of kinematics,

Here, negative sign shows direction of force is upward.