Question

A 4.2 x 10^3 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 18.4 degrees with the horizontal. An average frictional force of 4.1 x 10^3 N impedes the car's motion so that the car's speed at the bottom of the driveway is 4.9 m/s.
the acceleration of gravity is 9.81 m/s^2 . What is the length of the driveway?

Answer 1

Answer on Question#37801 - Physics - Other

A $4.2 \times 10^{3} \mathrm{~kg}$ car accelerates from rest at the top of a driveway that is sloped at an angle of 18.4 degrees with the horizontal. An average frictional force of $4.1 \times 10^{3} \mathrm{~N}$ impedes the car's motion so that the car's speed at the bottom of the driveway is 4.9 m/s.

the acceleration of gravity is $9.81 \, \text{m/s}^2$. What is the length of the driveway?

Solution:

First, we can write the Newton's second law along the slope:

$$F_{\text{net}} = \text{ma}$$ $$F_{\text{net}} = (\text{mg})_{\text{slope}} - F_{\text{friction}} = \text{mg} \cdot \sin \alpha - F_{\text{friction}}$$

(2)in(1):

$$\text{mg} \cdot \sin \alpha - F_{\text{friction}} = \text{ma}$$

Acceleration of the car:

$$\begin{array}{l} a = \frac{\text{mg} \cdot \sin \alpha - F_{\text{friction}}}{\text{m}} = \text{g} \cdot \sin \alpha - \frac{F_{\text{friction}}}{\text{m}} = \\ = 9.81 \frac{\text{m}}{\text{s}^2} \cdot \sin 18.4{}^\circ - \frac{4.1 \cdot 10^3 \text{N}}{4.2 \cdot 10^3 \text{kg}} = 2.12 \frac{\text{m}}{\text{s}^2} \end{array}$$

Equation of motion for the car:

$$d = \frac{a t^2}{2}$$

Rate equation for the car $(V_1 = 4.9 \frac{\text{m}}{\text{s}})$:

$$V_1 = a t \Rightarrow t = \frac{V_1}{a}$$

(4)in(3):

$$d = \frac{a \left(\frac{V_1}{a}\right)^2}{2} = \frac{V_1^2}{2a} = \frac{\left(4.9 \frac{\text{m}}{\text{s}}\right)^2}{2 \cdot 2.12 \frac{\text{m}}{\text{s}^2}} = 5.7 \text{m}$$

Answer: length of the driveway is equal to $5.7\text{m}$.