More practice questions

4. Q4. A 4 kg block slides on a rough horizontal floor. Its speed falls from 6 m s^⁻¹ to 2 m s^⁻¹ over 8 m. Assume friction is constant and is the only force doing work; find its work and magnitude.

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   Gravity and the normal force are perpendicular to the motion and do no work, so friction’s work is the whole net work, equal to the change in kinetic energy:

   $$\begin{aligned} W_f &= \tfrac{1}{2}(4)(2^{2} - 6^{2}) \\ &= 2 \times (-32) = -64\ \text{J} \end{aligned}$$

   Friction acts against the motion over 8 m, so −f × 8 = −64, and f = 8 N.

5. Q5. A 50 kg sled starts from rest at the top of a smooth snow slope 20 m high. Find its speed at the bottom using the work-energy theorem. Take g = 9.8 m s⁻².

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   On a smooth slope the normal force is perpendicular to the motion and does no work, so only gravity does work. Along a slope of angle θ, gravity’s component is mg sin θ and the slope is h/sin θ long, so its work is mgh:

   $$\begin{aligned} \tfrac{1}{2}mv^{2} &= mgh \\ v &= \sqrt{2gh} \\ &= \sqrt{2 \times 9.8 \times 20} \\ &= \sqrt{392} \approx 19.8\ \text{m s}^{-1} \end{aligned}$$

6. Q6. Two cars, one of mass 1000 kg and one of mass 2000 kg, move at the same speed. The same braking force acts on each. Which stops in a shorter distance, and by what factor?

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   The stopping distance is mv²/(2F). With the same speed and force, it is proportional to the mass, so the 1000 kg car stops in half the distance of the 2000 kg car.