A0.05-kg car starts from rest at a height of 0.95 m. assuming no friction, what is the kinetic energy of the car when it reaches the bottom of the hill? (assume g = 9.81 m/s2.)

0.466 J

Explanation:

We are given that

Mass of car=m=0.05 kg

Initial velocity of care,u=0

Height of car from the bottom of hill=h=0.95 m

Potential energy of car=

Using the formula

Potential energy of car=

Loss of potential energy=Kinetic energy

Kinetic energy of the car when it reaches the bottom of the hill=0.466 J

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft