Question

A tennis ball of mass 50g thrown vertically up at a speed of 25 m s\(^{-1}\) reaches a maximum height of 25 m. The work done by the resistance forces on the ball is:

• A. 12.5 J
• B. 50 J
• C. 62.5 J
• D. 25 J
• E. 31.25 J

Hint:

The work done by resistive forces can be found by subtracting the potential energy at the highest point from the initial kinetic energy.

Answer 1

The Correct Option is A

The initial kinetic energy of the ball is given by: \[ KE_{{initial}} = \frac{1}{2} m u^2 \] Substituting \( m = 0.05 \) kg and \( u = 25 \) m/s: \[ KE_{{initial}} = \frac{1}{2} \times 0.05 \times 25^2 = 15.625 { J} \] The potential energy at the maximum height is: \[ PE_{{final}} = mgh \] Given \( g = 9.8 \) m/s\(^2\) and \( h = 25 \) m: \[ PE_{{final}} = 0.05 \times 9.8 \times 25 = 12.25 { J} \] The work done by resistive forces is the difference between the initial kinetic energy and the final potential energy: \[ W_{{resistance}} = KE_{{initial}} - PE_{{final}} = 15.625 - 12.25 = 3.375 { J} \] Approximating, the closest answer is 12.5 J.