Questions: The puck in Figure P11.46 has a mass of 0.120 kg. The distance of the puck from the center of rotation is originally 40.0 cm, and the puck is sliding with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck. (Suggestion: Consider the change of kinetic energy.)

Solution Steps

• Mass of the puck, \( m = 0.120 \, \text{kg} \)
• Initial distance from the center, \( R_i = 40.0 \, \text{cm} = 0.40 \, \text{m} \)
• Initial speed, \( v_i = 80.0 \, \text{cm/s} = 0.80 \, \text{m/s} \)
• Distance the string is pulled, \( \Delta R = 15.0 \, \text{cm} = 0.15 \, \text{m} \)
• Final distance from the center, \( R_f = R_i - \Delta R = 0.40 \, \text{m} - 0.15 \, \text{m} = 0.25 \, \text{m} \)

Since there is no external torque, angular momentum is conserved. \[ L_i = L_f \] \[ m R_i v_i = m R_f v_f \] \[ R_i v_i = R_f v_f \] \[ 0.40 \, \text{m} \times 0.80 \, \text{m/s} = 0.25 \, \text{m} \times v_f \] \[ v_f = \frac{0.40 \times 0.80}{0.25} \] \[ v_f = 1.28 \, \text{m/s} \]

Initial kinetic energy: \[ KE_i = \frac{1}{2} m v_i^2 \] \[ KE_i = \frac{1}{2} \times 0.120 \, \text{kg} \times (0.80 \, \text{m/s})^2 \] \[ KE_i = \frac{1}{2} \times 0.120 \times 0.64 \] \[ KE_i = 0.0384 \, \text{J} \]

Final kinetic energy: \[ KE_f = \frac{1}{2} m v_f^2 \] \[ KE_f = \frac{1}{2} \times 0.120 \, \text{kg} \times (1.28 \, \text{m/s})^2 \] \[ KE_f = \frac{1}{2} \times 0.120 \times 1.6384 \] \[ KE_f = 0.098304 \, \text{J} \]

The work done on the puck is the change in kinetic energy. \[ W = KE_f - KE_i \] \[ W = 0.098304 \, \text{J} - 0.0384 \, \text{J} \] \[ W = 0.059904 \, \text{J} \]

Final Answer