IGCSE Physics 0625: Conservation of Energy with Resistive Forces on a Snowy Slope

A wooden sledge loaded with firewood has a total mass of $20\text{ kg}$ . It is released from rest at the top of a snow-covered slope and slides down to the bottom, descending through a vertical height of $7.5\text{ m}$ while travelling a distance of $50\text{ m}$ along the slope. As it slides, friction between the sledge and the snow acts as a constant resistive force of $10\text{ N}$ along the direction of motion. Take gravitational field strength $g = 10\text{ N/kg}$ .

(a) Calculate the loss in gravitational potential energy of the sledge and its load as it descends the slope. [2]

(b) Calculate the work done against friction as the sledge travels the $50\text{ m}$ along the slope. [2]

Part (a): Loss in gravitational potential energy

Use $\Delta E_p = mg\Delta h$ with $m = 20\text{ kg}$ , $g = 10\text{ N/kg}$ and $\Delta h = 7.5\text{ m}$ (the vertical height, not the slope distance):

$\Delta E_p = 20 \times 10 \times 7.5$

$\Delta E_p = 1500\text{ J}$

Part (b): Work done against friction

Use $W = Fd$ with the resistive force $F = 10\text{ N}$ acting over the slope distance $d = 50\text{ m}$ :

$W = 10 \times 50$

$W = 500\text{ J}$

Part (c): Finding the speed at the bottom

By conservation of energy, the gravitational potential energy lost is shared between the kinetic energy gained and the energy transferred by doing work against friction:

$E_{p,\text{lost}} = E_k + W_{\text{friction}}$

$1500 = E_k + 500$

$E_k = 1500 - 500 = 1000\text{ J}$

Now use $E_k = \tfrac{1}{2}mv^2$ to find the speed:

$1000 = \tfrac{1}{2} \times 20 \times v^2$

$1000 = 10 \times v^2$

$v^2 = \frac{1000}{10} = 100$

$v = \sqrt{100} = 10\text{ m/s}$

Final answers

(a) Loss in gravitational potential energy $=$ $1500\text{ J}$
(b) Work done against friction $=$ $500\text{ J}$
(c) Speed at the bottom of the slope $=$ $10\text{ m/s}$