Question
Topic:
Given: A girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is $16 \mathrm{~V}$ and it can deliver a charge of $43 \mathrm{kC}$ when discharging completely from a full charge.
The maximum speed of the girl on a horizontal road is $7.0 \mathrm{~m} \mathrm{~s}^{-1}$ with energy from the battery alone. The maximum distance that the girl can travel under these conditions is $20 \mathrm{~km}$
Prove: that the time taken for the battery to discharge is about $3 \times 10^3 \mathrm{~s}$.
▶️Answer/Explanation
Solution:
$V=\frac{d}{t}=>t=\frac{d}{v} ; d=20 \mathrm{~km}=20000 \mathrm{~m}, v=7.0 \mathrm{~m} / \mathrm{s}: \quad V=\frac{20000}{7.0}=2.85 \times 10^3 \approx 3 \times 10^3 \mathrm{~s}$
Question
Topic:
Show: that the average power output of the battery is about $240 \mathrm{~W}$.
▶️Answer/Explanation
Solution:
Using the formula $P = \frac{W}{t}$ and the equation for the voltage $V = \frac{W}{Q}$ to obtain $W = VQ$. Plugging in the given values, we get:
$$
P = \frac{W}{t} = \frac{VQ}{t} = \frac{(16 \mathrm{~V})(43 \times 10^3 \mathrm{~C})}{2.85 \times 10^3 \mathrm{~s}} \approx 241 \mathrm{~W} \approx 240 \mathrm{~W}
$$
So the average power output of the battery is about $240 \mathrm{~W}$, as previously calculated. Well done!
Question
Topic:
Given: Friction and air resistance act on the bicycle and the girl when they move. Assume that all the energy is transferred from the battery to the electric motor.
Calculate: the total average resistive force that acts on the bicycle and the girl.
▶️Answer/Explanation
Solution:
$F_{\text {res }}=F_{\text {engine: }}$
Power $=\mathrm{Fv} \Rightarrow \mathrm{F}=\frac{\text { power }}{\mathrm{v}}=\frac{241}{7.0}=34 \mathrm{~N}$
Question
Topic:
Given: The bicycle and the girl have a total mass of $66 \mathrm{~kg}$. The girl rides up a slope that is at an angle of $3.0^{\circ}$ to the horizontal.
Calculate: the component of weight for the bicycle and girl acting down the slope.
▶️Answer/Explanation
Solution:
The weight of the bicycle and the girl is given by:
$$
mg = (66\text{ kg})(9.81\text{ m/s}^2) \approx 648\text{ N}
$$
The component of weight acting down the slope is given by:
$$
mg\sin\theta = (648\text{ N})\sin(3.0^\circ) \approx 33.7\text{ N}
$$
where $\theta = 3.0^\circ$ is the angle of the slope with respect to the horizontal.
Question
Topic:
Given:The battery continues to give an output power of $240 \mathrm{~W}$. Assume that the resistive forces are the same as in (a)(iii).
Calculate: the maximum speed of the bicycle and the girl up the slope.
▶️Answer/Explanation
Solution:
Sum of backward forces $=F_{\text {res }}+F_{\mathrm{mg}}=34+34=68 \mathrm{~N}$
$$
\text { power }=\mathrm{Fv} \Rightarrow \mathrm{v}=\frac{\text { power }}{\mathrm{F}}=\frac{240 \mathrm{~W}}{68 \mathrm{~N}}=3.5 \mathrm{~m} / \mathrm{s}
$$
Question
Topic:
Discuss:On another journey up the slope, the girl carries an additional mass. Explain whether carrying this mass will change the maximum distance that the bicycle can travel along the slope.
▶️Answer/Explanation
Solution:
When the girl carries an additional mass, the total mass of the bicycle and the girl increases, which means that more work is required to lift the combined mass against the force of gravity as they move up the slope. This means that the initial potential energy of the system also increases, and the maximum distance that the bicycle can travel up the slope decreases, since more work is required to lift the increased mass to a certain height.
The opposing/resistive force and air resistance may also increase with the additional mass, but these factors alone would not be sufficient to explain why the maximum distance decreases. The key factor is the increased mass requiring more work to move up the hill.