A900-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

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.

increase

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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