Mathematics 9709/42 — May/June 2017

Walkthrough

Start by listing what is known: the tension is $20\text{ N}$, it acts at angle $\theta$ to the horizontal, the block moves at $1.5\text{ m s}^{-1}$ for $12\text{ s}$, and the work done by the tension is $50\text{ J}$.

The block moves horizontally, so only the horizontal component of the tension does work. That component is $20\cos\theta$. The vertical component $20\sin\theta$ is perpendicular to the displacement and does no work.

The distance travelled is

$$d = vt = 1.5 \times 12 = 18\text{ m}.$$

Using the work formula for a constant force acting at an angle,

$$W = Fd\cos\theta = (20)(18)\cos\theta = 360\cos\theta.$$

Set this equal to the given work:

$$50 = 360\cos\theta \quad \Rightarrow \quad \cos\theta = \frac{5}{36}.$$

Then

$$\theta = \cos^{-1}\left(\frac{5}{36}\right) = 82.0^{\circ}.$$

An equivalent method uses power. The power supplied by the tension is

$$P = \frac{W}{t} = \frac{50}{12} = \frac{25}{6}\text{ W}.$$

Since $P = Fv\cos\theta$,

$$\frac{25}{6} = (20)(1.5)\cos\theta = 30\cos\theta,$$

so $\cos\theta = \frac{25}{180} = \frac{5}{36}$, giving the same answer.

Key Takeaways

• Work done by a constant force at an angle to the displacement is $W = Fd\cos\theta$.
• Only the component of force in the direction of motion does work.
• Distance can be found from constant speed and time using $d = vt$.
• Power is the rate of doing work, and for a force at angle $\theta$ to the velocity, $P = Fv\cos\theta$.
• Inverse cosine is used to recover an angle from its cosine.

Common Mistakes

• Using the full $20\text{ N}$ without multiplying by $\cos\theta$.
• Using the time $12\text{ s}$ as the distance instead of computing $1.5 \times 12 = 18\text{ m}$.
• Treating $360\cos\theta$ as the answer for work, rather than setting it equal to $50\text{ J}$.
• Having the calculator in radians instead of degrees.
• Using the vertical component $20\sin\theta$ in the work formula; it does no work here.
• In the power method, forgetting the $\cos\theta$ factor in $P = Fv\cos\theta$.

Things to Be Careful About

• The angle is measured to the horizontal, so the horizontal component is $20\cos\theta$.
• Keep units consistent: newtons, metres, seconds, joules, watts.
• Constant speed means zero net work, but the tension itself can still do positive work while friction does negative work.
• The mark scheme requires the method to be shown: award B1 for $W = Fd\cos\theta$ (or $P = W/t$), M1 for substituting $W=50$ and solving, and A1 for $\theta = 82.0^{\circ}$.