Question 22

Draw the free-body diagram first. Four forces act on the cart: the push $F_p$ up the slope, the weight $F_g$ straight down, the normal force $F_N$ perpendicular to the slope and friction $F_f$ down the slope, because the cart is pushed up. Arrowheads and labels on all four.

Nothing moves perpendicular to the slope, so the normal force balances the component of the weight into the slope. Write that recognition down first, $F_N=F_\perp$, as the guide does.

$$\begin{aligned}F_N&=F_\perp=mg\cos\theta\\&=5\times9.8\times\cos30^\circ\\&=42.44\text{ N}\end{aligned}$$

The cart is moving, so use kinetic friction. Here $\mu_k=\mu_s$, so you never have to choose between them.

$$\begin{aligned}F_f&=\mu_kF_N\\&=0.58\times42.44\\&=24.62\text{ N}\end{aligned}$$

Keeping full precision gives 24.61 N. Both are inside the accepted range.

The component of the weight along the slope also points down the slope.

$$\begin{aligned}F_\parallel&=mg\sin\theta\\&=5\times9.8\times\sin30^\circ\\&=24.5\text{ N}\end{aligned}$$

The guide labels this the horizontal component of the weight. It is the component parallel to the slope, which is what its symbol $F_\parallel$ says.

Add the forces along the slope, taking up the slope as positive. The push is the only force pointing up it.

$$\begin{aligned}F_{net}&=F_p-F_\parallel-F_f\\&=60-24.5-24.62\\&=10.88\text{ N}\end{aligned}$$

The guide writes the push as $F_H$ in this line and as $F_p$ in its diagram notes. Use one symbol and stick to it.

Newton’s second law turns the net force into an acceleration.

$$\begin{aligned}a&=\frac{F_{net}}{m}=\frac{10.88}{5}\\&=2.18\text{ m s}^{-2}\end{aligned}$$

This mark is for choosing the method, so show the choice. List what you know: $u=0$ from “initially at rest”, $s=3$ m and $a=2.18$ m s⁻². You want $t$ and you do not know $v$, so use the motion equation with no $v$ in it.

$$\begin{aligned}s&=ut+\tfrac{1}{2}at^2\\3&=0\times t+\tfrac{1}{2}(2.18)t^2\\6&=2.18\,t^2\\t^2&=\frac{6}{2.18}\end{aligned}$$