Mathematics 9709/43 — May/June 2017

Walkthrough

The wheelbarrow moves horizontally, but the pushing force is $20^\circ$ below the horizontal. Work is done only by the component of the force in the direction of motion, which is $35\cos 20^\circ$. Multiplying by the distance $12\text{ m}$ gives the work done by the man. The vertical component of the push does no work because there is no vertical displacement.

Key Takeaways

Work done by a constant force is the product of the component of force along the displacement and the distance moved: $W = Fd\cos\theta$. Work is a scalar quantity measured in joules.

Common Mistakes

Using the full $35\text{ N}$ instead of its horizontal component; using $\sin 20^\circ$ instead of $\cos 20^\circ$; forgetting to multiply by the distance $12\text{ m}$; omitting units.

Things to Be Careful About

The angle is measured between the force and the horizontal displacement, so $\cos 20^\circ$ is correct even though the force is below the horizontal. Give the answer to 3 significant figures as $395\text{ J}$.

Walkthrough

First calculate the work done by the man in the same way as part (i): $35\cos 20^\circ \times 12 \approx 394.7\text{ J}$. The resistance of $15\text{ N}$ acts opposite to the motion over $12\text{ m}$, so it removes $15 \times 12 = 180\text{ J}$ of energy. The net work done on the wheelbarrow is therefore $394.7 - 180 = 214.7\text{ J}$. By the work-energy principle, this net work equals the gain in kinetic energy. Since the wheelbarrow starts from rest, its final kinetic energy is $\frac{1}{2}(25)v^2 = 12.5v^2$. Solving $12.5v^2 = 214.7$ gives $v \approx 4.14\text{ m s}^{-1}$.

Alternatively, use Newton's second law. The horizontal component of the push is $35\cos 20^\circ \approx 32.9\text{ N}$, so the resultant force is $32.9 - 15 = 17.9\text{ N}$. Then $a = \frac{17.9}{25} \approx 0.716\text{ m s}^{-2}$, and $v^2 = 2as = 2(0.716)(12) \approx 17.2$, giving $v \approx 4.14\text{ m s}^{-1}$.

Key Takeaways

The work-energy principle states that the total work done by all forces equals the change in kinetic energy. Work done against a resistance is energy removed from the system. Kinetic energy is $\frac{1}{2}mv^2$.

Common Mistakes

Forgetting to include the work done against resistance; using the full $35\text{ N}$ force rather than its horizontal component; using $m$ incorrectly in $\frac{1}{2}mv^2$; forgetting that the wheelbarrow starts from rest so initial KE is zero.

Things to Be Careful About

Work done by the man is positive, while work done by the resistance is negative; the net work equals the KE gain. In the alternative Newton's law method, use the horizontal component of the push and remember the acceleration is constant, so $v^2 = u^2 + 2as$ applies with $u=0$. Give final speed to 3 significant figures: $4.14\text{ m s}^{-1}$.