Suppose a car approaches a hill and has an initial speed of 112 km/h at the bott

Question

Suppose a car approaches a hill and has an initial speed of 112 km/h at the bottom of the hill. The driver takes her foot off of the gas pedal and allows the car to coast up the hill. If the car has the initial speed stated at a height of h = 0, how high (in m) can the car coast up a hill if work done by friction is negligible? If, in actuality, a 750-kg car with an initial speed of 112 km/h is observed to coast up a hill and stops at a height 21 m above its starting point, how much thermal energy was generated by friction in J? What is the magnitude of the average force in Newtons of friction if the hill has a slope 22 degree above the horizontal?

Explanation / Answer

using law of conservasion of energy

energy at the bottom = energy at the top

0.5*m*u^2 = m*g*h
ial speed of the car = 112*(5/18) = 31.12 m/sec

u is the init

m cancels

h = 0.5*u^2/g = (0.5*31.12^2)/9.81 = 49.4 m

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b) Work done by friction = change in mechanical energy

W_f = m*g*h - (0.5*m*v^2) = (750*9.81*21)-(0.5*750*31.12^2) = -208663 J

c) W_gravity +W_friction = chage in KE

(m*g*sin(2.2) + f_k)*S = 0.5*m*v^2

S = h/sin(2.2) = 21/sin(2.2) = 547 m

then

((750*9.8*sin(2.2)) + (f_k))*547 = 0.5*750*31.12^2

f_k = 381.8 N

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