Question

A 900-kg car initially at rest rolls 50 m down a hill inclined at an angle of 5.0°. A 400-N effective friction force opposes its motion.
A. What distance will it travel on a similar horizontal surface (assume the friction force is the same) at the bottom of the hill?
B. If the friction force remains the same will the distance decrease or increase if the car's mass is 1800 kg?
O will decrease
O will increase
O will remain the same

Answer 1

Answer:

d = 146.2 m

increase

Explanation:

Using Energy Balance:

U_i + W =  U_f

K_i + U_gi + W = K_f + U_fg

0 + m*g*h + F*d = 0.5*m*v_f^2

(900)*(9.81)*(50 sin (5)) + 400*50 = 0.5*900*v_f^2

v_f^2 = 129.944

v_f = 11.4 m/s at the bottom of the hill

U_i + W =  U_f

K_i + U_gi + W = K_f + U_fg

0.5*m*v_i^2 - F*d = 0

0.5*900*129.944 = 400*d

d = 146.19 m

part b

Heavier car will gain greater momentum at the bottom of the hill and attain a higher velocity at the bottom; hence, the distance will increase.

Answer: increase

Answer 2

(a) The distance traveled on a horizontal surface of similar friction is 146.2 m.

(b) If the car mass is 1800 kg, the distance traveled by the car will double because of increase in kinetic energy of the car.

The given parameters;

• mass of the car, = 900 kg
• distance down the hill, d = 50 m
• angle of inclination, θ = 5⁰
• frictional force, F = 400 N

Apply the principle of conservation of energy to determine the final speed of the car;

$mgsin\theta (d) + fd = \frac{1}{2} mv^2\\\\(900\times 9.8)sin(5)\times 50 \ + \ (400 \times 50) = \frac{1}{2} \times 900\times v^2\\\\v^2 = 129.86\\\\v = \sqrt{129.86} \\\\v = 11.4 \ m/s$

The distance traveled on a horizontal surface of similar friction is calculated as follows;

$fd = \frac{1}{2} mv^2\\\\400d = (0.5 \times 900 \times 11.4^2)\\\\400d = 58482\\\\d = \frac{58482}{400} \\\\d = 146.2 \ m$

If the car mass is 1800 kg, the distance traveled by the car will double because of increase in kinetic energy of the car.

Learn more here:https://brainly.com/question/14376066