Question

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story.
If a 1500 kg car is driven in sand at 5.0 m/s , the coefficient of rolling friction is 0.06. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s ?
c. If the car gets 15 miles per gallon when driving on sand, what is the car's efficiency? One gallon of gasoline contains 1.4 x 10^8 J of chemical energy, one mile is 1609 m.

Answer 1

Answer:

a) $F = 882.63\,N$, b) $\dot W= 4413.15\,W$, c) $\eta = 15.216\,\%$.

Explanation:

a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:

$\Sigma F_{x} = F - \mu_{r}\cdot N = 0$

$\Sigma F_{y} = N - m\cdot g = 0$

After some algebraic handling, the following expression for the propulsion force is constructed:

$F = \mu_{r}\cdot m \cdot g$

$F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 882.63\,N$

b) The power require to move the car at a speed of 5 meters per second is:

$\dot W = F\cdot v$

$\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )$

$\dot W= 4413.15\,W$

c) The efficiency of the car is:

$\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%$

$\eta = 15.216\,\%$