To determine the final velocity of the hockey puck, we can apply the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy.

Step 1: Calculate the initial kinetic energy (KE₁)

The initial kinetic energy $KE_1$ is given by: $KE_1 = \frac{1}{2} m v_1^2$ Where:

• $m = 0.20 \, \text{kg}$ (mass of the puck)
• $v_1 = 5.0 \, \text{m/s}$ (initial velocity)

$KE_1 = \frac{1}{2} (0.20) (5.0)^2 = \frac{1}{2} (0.20) (25) = 2.5 \, \text{J}$

Step 2: Calculate the work done by the force

The work $W$ done by the force is: $W = F \cdot d$ Where:

• $F = 7.0 \, \text{N}$ (force applied)
• $d = 8.0 \, \text{m}$ (distance over which the force is applied)

$W = 7.0 \times 8.0 = 56 \, \text{J}$

Step 3: Calculate the total kinetic energy (KE₂)

The total kinetic energy after the force is applied is: $KE_2 = KE_1 + W$ $KE_2 = 2.5 + 56 = 58.5 \, \text{J}$

Step 4: Calculate the final velocity

The final kinetic energy is related to the final velocity $v_2$ by: $KE_2 = \frac{1}{2} m v_2^2$ Solving for $v_2$: $v_2 = \sqrt{\frac{2 KE_2}{m}} = \sqrt{\frac{2 \times 58.5}{0.20}} = \sqrt{585} = 24.2 \, \text{m/s}$

Thus, the final velocity of the hockey puck is 24.2 m/s.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What would happen to the final velocity if the mass of the puck was doubled?
2. How would the final velocity change if the applied force was halved?
3. How much work would need to be done to bring the puck to rest after it reaches its final velocity?
4. What would the final velocity be if friction were not negligible?
5. How does the final velocity change if the initial velocity is increased?

Tip: Remember that work and energy concepts are closely related to force and motion. Understanding the work-energy principle can simplify many physics problems involving varying forces and distances.