1. **Problem statement:** A 5.0 kg block moves up a 30° inclined plane with initial speed 8.0 m/s and stops after 3.0 m. Calculate: a) Change in kinetic energy b) Change in potential energy c) Frictional force d) Coefficient of kinetic friction 2. **Formulas and rules:** - Kinetic energy: $KE = \frac{1}{2}mv^2$ - Potential energy change: $\Delta PE = mg\Delta h$ - Frictional force: $f_k = \mu_k N$ - Normal force on incline: $N = mg\cos\theta$ - Work-energy principle: Work done by friction and gravity changes kinetic energy 3. **a) Change in kinetic energy:** Initial kinetic energy: $KE_i = \frac{1}{2} \times 5.0 \times 8.0^2 = 160$ J Final kinetic energy: $KE_f = 0$ J (block stops) Change: $\Delta KE = KE_f - KE_i = 0 - 160 = -160$ J 4. **b) Change in potential energy:** Height change: $\Delta h = 3.0 \times \sin 30^\circ = 3.0 \times 0.5 = 1.5$ m Change in potential energy: $\Delta PE = mg\Delta h = 5.0 \times 9.8 \times 1.5 = 73.5$ J 5. **c) Frictional force:** Total work done by friction and gravity equals change in kinetic energy: $$ W = f_k \times 3.0 + (-mg\sin 30^\circ) \times 3.0 = \Delta KE $$ Let friction force $f_k$ act opposite motion (uphill positive, friction downhill negative): $$ -f_k \times 3.0 - mg\sin 30^\circ \times 3.0 = -160 $$ Calculate gravity component: $$ mg\sin 30^\circ = 5.0 \times 9.8 \times 0.5 = 24.5 \text{ N} $$ Substitute: $$ -3.0 f_k - 3.0 \times 24.5 = -160 $$ $$ -3.0 f_k - 73.5 = -160 $$ $$ -3.0 f_k = -160 + 73.5 = -86.5 $$ $$ f_k = \frac{86.5}{3.0} = 28.83 \text{ N} $$ 6. **d) Coefficient of kinetic friction:** Normal force: $$ N = mg\cos 30^\circ = 5.0 \times 9.8 \times 0.866 = 42.4 \text{ N} $$ Coefficient: $$ \mu_k = \frac{f_k}{N} = \frac{28.83}{42.4} = 0.68 $$ **Final answers:** a) $\Delta KE = -160$ J b) $\Delta PE = 73.5$ J c) $f_k = 28.8$ N d) $\mu_k = 0.678$