91影视

Newton鈥檚 second law of motion states that the acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.

Mathematically,

$a鈭�\frac{F_{\mathrm{net}}}{m}$

The mass of the car is $m=75\mathrm{kg}$.

The initial velocity of the car is $u=0$.

The final velocity of the car is $v=21\mathrm{m}/\mathrm{s}$.

The time taken is $t=12.5\mathrm{s}$.

A free body diagram of the car climbing up the hill is drawn below.

The forces acting on the moving car are:

• Normal force, N exerted upward on the car
• Weight, W of the car acting downward
• Force, F exerted to push the car against the road

Using the following kinematic equation of motion to determine the acceleration of the car on level ground, you get:

$\begin{array}{c}v=u+at\\ a=\frac{v-u}{t}\\ =\frac{21\mathrm{m}/\mathrm{s}-0}{12.5\mathrm{s}}\\ =1.68\mathrm{m}/{\mathrm{s}}^2\end{array}$

From Newton鈥檚 second law of motion, the force pushing the car forward to have acceleration is given by:

$\begin{array}{c}F=ma\\ =\left(920\mathrm{kg}\right)\left(1.68\mathrm{m}/{\mathrm{s}}^2\right)\\ 鈮�1546\mathrm{N}\end{array}$

It is given that the car does not slow down while climbing the hill. Assume that the car has a constant speed with zero acceleration. The force, F, will be equal to the magnitude of the weight component directed down the hill.

From Newton鈥檚 second law of motion,

$\begin{array}{c}F=mg\sin 胃\\ \sin 胃=\frac{F}{mg}\\ 胃={\sin }^{-1}\left(\frac{F}{mg}\right)\end{array}$

Substitute the numerical values in the above expression for the maximum angle of the hill.

$\begin{array}{c}胃={\sin }^{-1}\left(\frac{1546\mathrm{N}}{\left(920\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\right)\\ =9.9^{\mathrm{o}}\end{array}$

Thus, the maximum angle for the hill is $9.9^{\mathrm{o}}$.