3. A block of mass 2 kg slides along a horizontal surface that has negligible friction except for one tection, as shown. The block arrives at the rough section with a speed of \( 4 \mathrm{~m} / \mathrm{s} \) and leaves it 0.5 s later with a speed of \( 1 \mathrm{~m} / \mathrm{s} \). What is the work done by frictional force exerted on the block?

To find the work done by the frictional force, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. 1. **Calculate initial and final kinetic energy**: - Initial kinetic energy (KE_initial) = \(\frac{1}{2} m v_i^2 = \frac{1}{2} \times 2 \, \text{kg} \times (4 \, \text{m/s})^2 = 16 \, \text{J}\) - Final kinetic energy (KE_final) = \(\frac{1}{2} m v_f^2 = \frac{1}{2} \times 2 \, \text{kg} \times (1 \, \text{m/s})^2 = 1 \, \text{J}\) 2. **Calculate the change in kinetic energy**: - Change in kinetic energy = KE_final - KE_initial = \(1 \, \text{J} - 16 \, \text{J} = -15 \, \text{J}\) 3. **Work done by the frictional force**: - Since the work done by friction is equal to the change in kinetic energy, the work done by the frictional force is \(-15 \, \text{J}\). Thus, the work done by the frictional force exerted on the block is \(-15 \, \text{J}\).