Energy, work and power — practice questions

We will use the work-energy principle. Let the initial position be at the top of the slope and the final position be at the bottom.

1. Calculate Initial Energy: The box is at rest, so $KE_i = 0$. The vertical height $h = 5 \sin 30^\circ = 2.5$ m. $GPE_i = mgh = 10 \times 9.8 \times 2.5 = 245$ J. Total initial energy $E_i = 245$ J.
2. Calculate Final Energy: At the bottom, we define the height as zero, so $GPE_f = 0$. Let the final speed be $v$. $KE_f = \frac{1}{2}mv^2 = \frac{1}{2}(10)v^2 = 5v^2$. Total final energy $E_f = 5v^2$.
3. Calculate Work Done Against Friction: First, find the normal reaction force, $R$. Resolving perpendicular to the slope: $R = mg \cos 30^\circ = 10 \times 9.8 \times \frac{\sqrt{3}}{2} = 49\sqrt{3}$ N. The frictional force is $F_f = \mu R = 0.2 \times 49\sqrt{3} \approx 16.974$ N. The work done against friction over the distance of the slope (5 m) is $W_f = F_f \times d = 16.974 \times 5 \approx 84.87$ J. This is energy lost from the system.
4. Apply the Energy Principle: Initial Energy = Final Energy + Work Done Against Friction $E_i = E_f + W_f$ $245 = 5v^2 + 84.87$ $5v^2 = 245 - 84.87 = 160.13$ $v^2 = \frac{160.13}{5} = 32.026$ $v = \sqrt{32.026} \approx 5.66$ m/s (3 s.f.).

(a) Finding acceleration at v = 20 m/s:

1. Convert Power: Power $P = 30 \text{ kW} = 30000$ W.
2. Find Driving Force: Use $P = F_D v$, where $F_D$ is the driving force. $30000 = F_D \times 20$ $F_D = \frac{30000}{20} = 1500$ N.
3. Find Resultant Force: The road is horizontal. Resultant Force = Driving Force - Resistance. $F_{net} = 1500 - 500 = 1000$ N.
4. Apply Newton's Second Law: $F_{net} = ma$. $1000 = 1200a$ $a = \frac{1000}{1200} = \frac{5}{6} \approx 0.833$ m/s² (3 s.f.).

(b) Finding maximum speed:

1. Condition for Maximum Speed: Maximum speed ($v_{max}$) is reached when acceleration is 0. This means the resultant force is 0.
2. Equate Forces: At maximum speed, Driving Force = Resistance. $F_D = 500$ N.
3. Use Power Formula: Use $P = F_D v_{max}$ with the constant power output. $30000 = 500 \times v_{max}$ $v_{max} = \frac{30000}{500} = 60$ m/s.