contestada

A 1800 kg car is parked at the top of a hill 4.7 m high.a. What is the gravitational potential energy of the car?b. The car's brakes fail and it rolls down the hill. At the bottom of the hill, what isthe potential energy of the car?c. At the bottom of the hill, what is the kinetic energy of the car?d. What is the velocity of the car when it reaches the bottom of the hill?

Respuesta :

mass = m = 1800 kg

height = h = 4.7 m (top of the hill)

a) gravitational potential energy = m x g x h

PE = 1800 x 9.8m/s^2 x 4.7m = 82,908 J

b) bottom of the hill height = 0

PE = 1800 x 9.8 x 0 = 0 J

c) Kinetic energy = 1/2 x m x v^2

At the bottom of the hill PE converts into KE

82,908 J

d) Kinetic energy = 1/2 x m x v^2

82,908 =1/2 m v^2

82,908= 1/2 * 1800* v^2

82,908/ (1/2*1800) = v^2

√[82,908/ (1/2*1800)] = v

v = 9.6 m/s