Mathematics 9709/42 — October/November 2023

Walkthrough

This is a work-energy problem. The block starts at the top of the plane with both kinetic energy and gravitational potential energy, and it reaches the bottom with only kinetic energy. Since resistance acts, mechanical energy is not conserved; the missing energy is the work done against the resistance.

Step 1: Calculate the kinetic energy at the top using $\frac{1}{2}mv^2$:

$$\frac{1}{2}(15)(2^2)=30\text{ J}$$

Step 2: Calculate the gravitational potential energy at the top relative to the bottom. The vertical height is $1.6\text{ m}$, so:

$$mgh = 15(10)(1.6)=240\text{ J}$$

Take $g=10\text{ m s}^{-2}$, as used by the mark scheme.

Step 3: Calculate the kinetic energy at the bottom:

$$\frac{1}{2}(15)(4^2)=120\text{ J}$$

Step 4: Write the energy balance:

$$30+240=120+W$$

where $W$ is the work done against resistance. Solving:

$$W = 270 - 120 = 150\text{ J}$$

The block's potential energy provides both the increase in kinetic energy from $30\text{ J}$ to $120\text{ J}$ and the $150\text{ J}$ of work done against resistance.

Key Takeaways

The question tests the work-energy principle, gravitational potential energy, and kinetic energy. A useful idea is to treat all mechanical energy at the start as the total available to be converted into final mechanical energy plus any work done by non-conservative forces such as resistance.

Common Mistakes

• Using the change in speed $(4-2)$ in a single kinetic-energy term. The mark scheme explicitly rejects $\frac{1}{2} \times 15 \times (4-2)^2$; each state must use its own speed.
• Omitting the initial kinetic energy or the potential energy term.
• Writing $W$ as a force multiplied by a numerical distance in the work-energy equation without a clear displacement; the work against resistance is the energy term found from the balance.
• Using the sloping distance instead of the vertical height for potential energy. Gravitational potential energy always uses vertical height.

Things to Be Careful About

The standard value used in this question is $g=10\text{ m s}^{-2}$, so $mg = 150\text{ N}$ and the potential energy is $240\text{ J}$.

Be careful with signs. The work done against resistance is positive energy removed from the block, so it appears on the right-hand side of the energy equation.

If you use the alternative Newton's laws and suvat method, the distance along the slope must be written as $\frac{1.6}{\sin\theta}$, not as $1.6$. The energy method avoids this by using the vertical height directly.

At the bottom, $\frac{1}{2}(15)(4^2) = 120\text{ J}$; this is more than the initial kinetic energy, so the increase comes from the loss of potential energy.