Towing a Car
A damaged 1200-kg car is being towed by a truck. Neglecting the friction, air drag, and rolling resistance, determine the extra power required (a) for constant velocity on a level road, (b) for constant velocity of 50 km/h on a 30 degree (from horizontal) uphill road, and (c) to accelerate on a level road from stop to 90 km/h in 12 seconds.
Hint
The power due to potential energy:
$$$\dot{W}_{g}=mg(z_{2}-z_{1})/\Delta t$$$
where
$$m$$
is the mass,
$$g$$
is the acceleration due to gravity,
$$\Delta z$$
is the change in height, and
$$\Delta t$$
is the change in time.
Hint 2
The power from kinetic energy:
$$$\dot{W}_{a}=\frac{1}{2}m(V_{2}^{2}-V_{1}^{2})/\Delta t$$$
where
$$m$$
is the mass,
$$V$$
is the velocity, and
$$\Delta t$$
is the change in time.
The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is:
$$$\dot{W}_{total}=\dot{W}_{a}+\dot{W}_{g}$$$
where
$$\dot{W}_a$$
is the power from kinetic energy, and
$$\dot{W}_g$$
is the power from potential energy.
(a) Zero.
(b)
$$\dot{W}_{a}=0$$
. Thus,
$$$\dot{W}_{total}=\dot{W}_{g}=mg(z_{2}-z_{1})/\Delta t=mg\frac{\Delta z}{\Delta t}=mgV_{2}=mgVsin(30^{\circ})$$$
where
$$m$$
is the mass,
$$g$$
is the acceleration due to gravity,
$$\Delta z$$
is the change in height,
$$\Delta t$$
is the change in time,
$$V$$
is the velocity. Thus, the extra power on an upward hill is:
$$$=(1200kg)(9.81m/s^2)(\frac{50,000m}{3600s})(\frac{1kJ/kg}{1000m^2/s^2})(0.5)=81.7\:kW$$$
(c)
$$\dot{W}_{g}=0$$
. Thus,
$$$\dot{W}_{total}=\dot{W}_{a}=\frac{1}{2}m(V_{2}^{2}-V_{1}^{2})/\Delta t$$$
where
$$m$$
is the mass,
$$V$$
is the velocity, and
$$\Delta t$$
is the change in time. Thus, the extra power to accelerate from rest is:
$$$=\frac{1}{2}(1200kg)[(\frac{90,000m}{3600s})^{2}-0](\frac{1kJ/kg}{1000m^2/s^2})/(12s)=31.3\:kW$$$
(a) Zero
(b) 81.7 kW
(c) 31.3 kW