A student releases a toy cart of mass 0.80 kg from rest at the top of a smooth track. The cart rolls down through a vertical height of 1.25 m before reaching a horizontal section at the bottom. Friction and air resistance are negligible. Take gravitational field strength g = 10…

Answer: ((a)) 10 J ((b)) Kinetic energy store ((c)) 5.0 m/s. Part (a): Loss in gravitational potential energy Use Ep = mg h with m = 0.80 kg, g = 10 N/kg and h = 1.25 m: Ep = 0.80 × 10 × 1.25 Ep = 10 J Part (b): Where the energy goes As the cart speeds up, this energy is transferred to the cart's ki…

Energy, Work and Power: Question 3

Syllabus 1.7.1

Structured Extended 6 marks

A student releases a toy cart of mass $0.80\text{ kg}$ from rest at the top of a smooth track. The cart rolls down through a vertical height of $1.25\text{ m}$ before reaching a horizontal section at the bottom. Friction and air resistance are negligible. Take gravitational field strength $g = 10\text{ N/kg}$ .

(a) Calculate the loss in gravitational potential energy of the cart as it descends the $1.25\text{ m}$ . [2]

(b) State the name of the energy store that this lost gravitational potential energy is transferred to as the cart speeds up. [1]

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Part (a): Loss in gravitational potential energy

Use $\Delta E_p = mg\Delta h$ with $m = 0.80\text{ kg}$ , $g = 10\text{ N/kg}$ and $\Delta h = 1.25\text{ m}$ :

$\Delta E_p = 0.80 \times 10 \times 1.25$

$\Delta E_p = 10\text{ J}$

Part (b): Where the energy goes

As the cart speeds up, this energy is transferred to the cart’s kinetic energy store.

Part (c): Finding the speed at the bottom

Since friction and air resistance are negligible, conservation of energy means the loss in gravitational potential energy equals the gain in kinetic energy:

$E_k = \Delta E_p = 10\text{ J}$

Use $E_k = \tfrac{1}{2}mv^2$ and solve for $v$ :

$10 = \tfrac{1}{2} \times 0.80 \times v^2$

$10 = 0.40 \times v^2$

$v^2 = \frac{10}{0.40} = 25$

$v = \sqrt{25} = 5.0\text{ m/s}$

Final answers

(a) Loss in gravitational potential energy $=$ $10\text{ J}$
(b) Energy store gained $=$ kinetic energy store
(c) Speed at the bottom $=$ $5.0\text{ m/s}$

Common mistakes

Leaving out $g$ when calculating the gravitational potential energy lost, e.g. working out $0.80 \times 1.25 = 1.0\text{ J}$ instead of $0.80 \times 10 \times 1.25 = 10\text{ J}$ .
Naming the store gained in part (b) as chemical or elastic energy instead of the kinetic energy store.
Omitting the factor of $\tfrac{1}{2}$ in the kinetic energy equation, giving $v^2 = 10 \div 0.80 = 12.5$ and $v \approx 3.5\text{ m/s}$ instead of $v^2 = 25$ .
Forgetting to take the square root at the end, leaving $25$ as the final speed instead of $5.0\text{ m/s}$ .
Assuming that some of the gravitational potential energy is used up rather than recognising that, with friction negligible, all of it is transferred to the kinetic energy store.