More practice questions

4. Q4. A 2 kg block is pressed against a vertical wall by a horizontal force F. The coefficient of static friction between the block and the wall is 0.5. What is the least F that stops the block from sliding down? Take g = 9.8 m s⁻².

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   Here the normal force is F itself, and static friction must hold up the weight: μ_s F ≥ mg. So

   $$F_{\min} = \frac{mg}{\mu_s} = \frac{19.6}{0.5} = 39.2\ \text{N}$$

5. Q5. A block sliding on a horizontal floor at 6 m s^⁻¹ comes to rest after 3 m. Find the coefficient of kinetic friction. Take g = 9.8 m s⁻².

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   The deceleration is

   $$a = \frac{v^{2}}{2s} = \frac{36}{6} = 6\ \text{m s}^{-2}$$

   and a = μ_k g, so μ_k = 6/9.8, about 0.612.

6. Q6. A spring with spring constant 200 N m^⁻¹ is stretched by 5 cm. Find the force it exerts, and its direction.

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   The size of the force is kx = 200 × 0.05, that is 10 N, directed back towards the spring’s natural length (opposite to the stretch), as the minus sign in F = −kx shows.

7. Q7. Why does a machine run more easily when ball bearings are placed between its moving parts?

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   Ball bearings replace sliding between the parts by rolling of the balls. Rolling friction is far smaller than sliding friction (by two to three orders of magnitude for the same weight), so much less power is wasted as heat and the parts wear less.

8. Q8. A box rests on the floor of a truck. The coefficients of friction are μ_s = 0.3 and μ_k = 0.25. (a) What is the greatest acceleration of the truck for which the box does not slip? (b) If the truck accelerates at 4 m s^⁻², find the box’s acceleration relative to the ground and relative to the truck. Take g = 9.8 m s⁻².

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   (a) Static friction is the only horizontal force on the box, so

   $$\begin{aligned} a_{\max} &= \mu_s g = 0.3 \times 9.8 \\ &= 2.94\ \text{m s}^{-2} \end{aligned}$$

   (b) 4 m s^⁻² is more than this, so the box slips. Kinetic friction, acting forward, gives it

   $$\begin{aligned} a &= \mu_k g = 0.25 \times 9.8 \\ &= 2.45\ \text{m s}^{-2} \end{aligned}$$

   relative to the ground. Relative to the truck its acceleration is 2.45 − 4, that is −1.55 m s^⁻²: it slides backward towards the rear of the truck.