Question
The total mass of a cyclist and her bicycle is 75 kg. The cyclist ascends a straight hill of length 0.7 km inclined at 1.5Å to the horizontal. Her speed at the bottom of the hill is 10 \(ms^{-1}\) and at the top it is 5\( ms^{-1}\)
. There is a resistance to motion, and the work done against this resistance as the cyclist ascends the hill is 2000 J. The cyclist exerts a constant force of magnitude F N in the direction of motion. Find F.
▶️Answer/Explanation
The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy. We need to find the force exerted by the cyclist (F) to overcome resistance and ascend the hill.
– Initial velocity (v_initial) = 10 m/s
– Final velocity (v_final) = 5 m/s
– Mass (m) = 75 kg
– Distance (d) = 0.7 km = 700 m
– Angle of incline (θ) = 1.5 degrees
– Work done against resistance (W_resistance) = 2000 J
The initial kinetic energy (KE_initial) and final kinetic energy (KE_final) of the cyclist and her bicycle:
\(KE_initial = \frac{1}{2} \times m\times v_initial^2 = \frac{1}{2}\times 75 kg\times 10 m/s^2 = 3750 J\)
\(KE_final = \frac{1}{2}\times m \times \times v_final^2 = \frac{1}{2}\times 75 kg\times (5 m/s)^2 = 937.5 J\)
ΔKE = KE_final – KE_initial = 937.5 J – 3750 J = -2812.5 J
The negative sign indicates a decrease in kinetic energy.
\(W_resistance = 2000 J \)(given)
Net Work \((W_net) = ΔKE + W_resistance\)
\(W_net = -2812.5 J + 2000 J = -812.5 J\)
\(W_cyclist = F \times distance \times cos(θ)\)
\(W_cyclist =F \times 700 m \times cos(1.5^{\circ})\)
\(W_cyclist – W_resistance = W_net\)
\(F \times 700 m \times cos(1.5^{\circ})- 2000 J = -812.5 J\)
\(F \times 700 m \times cos(1.5^{\circ}) = -812.5 J + 2000 J\)
\(F \times 700 m \times cos(1.5^{\circ}) = 1187.5 J\)
\(F =\frac{118.75}{700\times \cos (1.5^{\circ})}\)
F ≈ 18.5 N
So, the force exerted by the cyclist is approximately 18.5 N.
Question
A tractor of mass 3700 kg is travelling along a straight horizontal road at a constant speed of 12 \(m s^{−1}\). The total resistance to motion is 1150 N.
(i) Find the power output of the tractor’s engine. The tractor comes to a hill inclined at 4Å above the horizontal. The power output is increased to 25 kW and the resistance to motion is unchanged.
(ii) Find the deceleration of the tractor at the instant it begins to climb the hill.
(iii) Find the constant speed that the tractor could maintain on the hill when working at this power.
▶️Answer/Explanation
(i) The power output of the tractor’s engine on the horizontal road is correctly calculated as:
\(P = \text{Resistance} \times \text{Velocity} = 1150 \, \text{N} \times 12 \, \text{m/s} = 13,800 \, \text{W}\)
So, the power output on the horizontal road is 13,800 W, which is equivalent to 13.8 kW.
(ii) The deceleration (negative acceleration) is correctly calculated using the work-energy principle:
\(F_{\text{driving}} = \frac{25000 \, \text{W}}{12 \, \text{m/s}}\)
\(a = \frac{F_{\text{driving}} – \text{Resistance} – \text{Weight component parallel to the hill}}{m} = \frac{25000/12 – 1150 – 3700g \sin(4^\circ)}{3700}\)
Calculating this gives \(a \approx -0.445 \, \text{m/s}^2\), indicating deceleration.
(iii) The constant speed on the hill is correctly calculated using the power formula:
\(\frac{25000 \, \text{W}}{v} = 1150 + 3700g \sin(4^\circ)\)
\(v = \frac{25000 \, \text{W}}{1150 + 3700g \sin(4^\circ)} \approx 6.70 \, \text{m/s}\)
So, the constant speed that the tractor can maintain on the hill when working at 25 kW is approximately 6.70 m/s.