1. **State the problem:** A puck of mass 0.45 kg is struck with an initial velocity of 9.5 m/s and slides on smooth ice for 2.0 s before hitting a sanded area with kinetic friction coefficient $\mu_k = 0.65$. The puck starts 50 m from the goal. We need to determine if the puck reaches the goal. 2. **Identify known values:** - Mass, $m = 0.45$ kg - Initial velocity, $v_0 = 9.5$ m/s - Time on smooth ice, $t_1 = 2.0$ s - Coefficient of kinetic friction on sanded area, $\mu_k = 0.65$ - Distance to goal, $d = 50$ m 3. **Calculate distance traveled on smooth ice:** Since the ice is smooth (frictionless), the puck moves at constant velocity: $$ d_1 = v_0 \times t_1 = 9.5 \times 2.0 = 19 \text{ m} $$ 4. **Calculate velocity when puck hits sanded area:** Velocity remains $v_1 = 9.5$ m/s because no friction on ice. 5. **Calculate distance remaining to goal after ice:** $$ d_2 = d - d_1 = 50 - 19 = 31 \text{ m} $$ 6. **Calculate deceleration due to kinetic friction on sanded area:** Friction force $F_f = \mu_k mg$ Acceleration (deceleration) $a = \frac{F_f}{m} = \mu_k g$ Using $g = 9.8$ m/s$^2$: $$ a = 0.65 \times 9.8 = 6.37 \text{ m/s}^2 $$ 7. **Calculate stopping distance on sanded area:** Using kinematic equation for stopping distance with initial velocity $v_1$ and acceleration $-a$: $$ v^2 = v_1^2 - 2 a d_{stop} $$ At stop, $v=0$, so: $$ 0 = v_1^2 - 2 a d_{stop} \implies d_{stop} = \frac{v_1^2}{2 a} = \frac{9.5^2}{2 \times 6.37} = \frac{90.25}{12.74} \approx 7.08 \text{ m} $$ 8. **Compare stopping distance to remaining distance:** The puck can only travel about 7.08 m on the sanded area before stopping, but it needs to cover 31 m. **Conclusion:** The puck does not reach the goal because it stops after about 7.08 m on the sanded area, which is less than the 31 m remaining. **Final answer:** No, the puck does not reach the goal.