Mathematics 9709/42 — May/June 2015

Walkthrough

We are given the initial speed, the constant acceleration, and the time interval. Since the acceleration is constant, the suvat equations apply. The most direct formula for displacement is $s = ut + \frac{1}{2}at^2$. Substitute $u = 0.3\text{ m s}^{-1}$, $a = 0.5\text{ m s}^{-2}$ and $t = 5\text{ s}$. The term $0.3 \times 5$ gives the distance the block would travel at constant speed, while $\frac{1}{2} \times 0.5 \times 5^2$ accounts for the extra distance gained from acceleration. Adding these gives $7.75\text{ m}$.

Key Takeaways

This part tests the application of the constant acceleration formula for displacement. It is important to recognise which suvat variables are given and to choose the formula containing exactly those variables and the required quantity.

Common Mistakes

A common error is to use $v^2 = u^2 + 2as$ immediately without first finding the final velocity, or to use the wrong time exponent in the $\frac{1}{2}at^2$ term. Another common mistake is forgetting to halve the $at^2$ term.

Things to Be Careful About

Check that all units are consistent: distances in metres, speed in metres per second, acceleration in metres per second squared and time in seconds. Also note that the tension and angle given in the question are not needed for part (i); the horizontal motion is fully described by the acceleration already provided.

Walkthrough

From part (i), the block travels $7.75\text{ m}$ during the first $5\text{ s}$. The pulling force is the tension $8\text{ N}$, but the string is inclined at $60^{\circ}$ above the horizontal. Therefore not all of the tension does work in the direction of motion; only the horizontal component $T\cos 60^{\circ}$ is parallel to the displacement. The work done is

Since $\cos 60^{\circ} = 0.5$, this gives $31\text{ J}$. The vertical component $T\sin 60^{\circ}$ does no work in the horizontal motion, so the formula $Fd\cos\theta$ automatically picks out only the component in the direction of displacement.

Key Takeaways

Work is a scalar but is calculated using the component of force in the direction of displacement. The formula $W = Fd\cos\theta$ handles this, where $\theta$ is the angle between the force vector and the displacement vector. This part also shows the importance of using the result from part (i) to answer a follow-up question.

Common Mistakes

A common mistake is to use the full $8\text{ N}$ without multiplying by $\cos 60^{\circ}$, which would give $62\text{ J}$ instead of $31\text{ J}$. Another common mistake is to use $\sin 60^{\circ}$ or to forget that the angle should be measured between the force and the displacement, which are the diagonal tension and the horizontal motion in this setup.

Things to Be Careful About

The angle $60^{\circ}$ is measured above the horizontal, and the displacement is horizontal, so the angle between the tension and the displacement is indeed $60^{\circ}$. Use $\cos 60^{\circ} = 0.5$. Work is measured in joules, so the final answer should be stated as $31\text{ J}$ rather than just $31$.