A 1 kg block is pushed 10 m up a 30° incline by a 10 N force parallel to the plane. If the coefficient of kinetic friction is 0.1, what are (1) the work done against gravity, (2) the work done against friction, and (3) the block’s acceleration up the incline?

What you are being asked to find

As the block moves up the incline, two forces oppose the motion along the plane: the component of gravity down the plane and kinetic friction. The “work done against” each one is the energy you must supply to overcome it over the 10 m displacement.

Work done against gravity (gain in gravitational potential energy)

The vertical height gained is $$h = d\sin\theta$$ So the work done against gravity is $$W_g = mgh = mg(d\sin\theta)$$ Substitute $m=1\ \text{kg}$, $g=9.8\ \text{m/s}^2$, $d=10\ \text{m}$, $\sin 30^\circ = 0.5$: $$W_g = (1)(9.8)(10\cdot 0.5)=49\ \text{J}$$

Work done against friction

The normal force on an incline is $$N = mg\cos\theta$$ Kinetic friction magnitude is $$f_k = \mu N = \mu mg\cos\theta$$ Work done against friction over distance $d$ is $$W_f = f_k d = (\mu mg\cos\theta)d$$ With $\mu=0.1$ and $\cos 30^\circ \approx 0.866$: $$f_k = (0.1)(1)(9.8)(0.866) \approx 0.8487\ \text{N}$$ $$W_f = (0.8487)(10) \approx 8.49\ \text{J}$$

Acceleration up the incline (Newton’s 2nd law along the plane)

Take “up the plane” as positive. The forces along the plane are:

• Push: $+F$
• Gravity component down the plane: $-mg\sin\theta$
• Friction down the plane (opposes upward motion): $-f_k$

So $$\Sigma F = F - mg\sin\theta - \mu mg\cos\theta = ma$$ Compute each term: $$mg\sin 30^\circ = (1)(9.8)(0.5)=4.9\ \text{N}$$ $$\mu mg\cos 30^\circ \approx 0.8487\ \text{N}$$ Net force: $$F_\text{net} = 10 - 4.9 - 0.8487 \approx 4.2513\ \text{N}$$ Acceleration: $$a = \frac{F_\text{net}}{m} \approx \frac{4.2513}{1} = 4.25\ \text{m/s}^2$$ The block accelerates upward because the applied force is larger than the downhill gravity component plus friction.