A player catches a ball of mass 200g moving with speed of 5m/s by lowering his hands to a distance of .5m. The average force exerted by ball on his hands will be

To solve this problem, we need to use the principles of physics, specifically the concept of work and energy.

Step 1: Identify the given values

• Mass (m) = 200g = 0.2 kg (converted from grams to kilograms)
• Initial speed (u) = 5 m/s
• Final speed (v) = 0 m/s (since the ball is caught and comes to rest)
• Distance (s) = 0.5 m

Step 2: Use the work-energy theorem The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the player's hands in stopping the ball is equal to the change in the ball's kinetic energy.

Step 3: Calculate the initial kinetic energy of the ball The kinetic energy (KE) of an object can be calculated using the formula KE = 1/2 * m * u^2. Substituting the given values, we get KE_initial = 1/2 * 0.2 kg * (5 m/s)^2 = 2.5 J.

Step 4: Calculate the final kinetic energy of the ball Since the ball comes to rest, its final kinetic energy is 0 J.

Step 5: Calculate the work done The work done (W) is equal to the change in kinetic energy, which is the final kinetic energy minus the initial kinetic energy. So, W = KE_final - KE_initial = 0 J - 2.5 J = -2.5 J. The negative sign indicates that the work is done against the motion of the ball.

Step 6: Calculate the average force Work done is also equal to the force (F) times the distance (s) over which the force is applied. So, we can rearrange the formula to solve for force: F = W / s = -2.5 J / 0.5 m = -5 N. The negative sign indicates that the force is exerted in the opposite direction of the ball's motion.

So, the average force exerted by the ball on the player's hands is 5 N.