Question

A 0.05-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.)

Answer 1

Answer:

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

$g=9.81 m/s^2$

Potential energy of car=$mgh$

Using the formula

Potential energy of car=$0.05\times 0.95\times 9.81=0.466 J$

Loss of potential energy=Kinetic energy

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

Answer 2

     Since there is no friction means no dissipative forces, so the mechanical energy is conserved. When the car reaches the zero point, all the mechanical energy, that is, the Home Energy Potecial Gravitational, will be converted into kinetic energy. Through this, we have:

$E_{c}=E_{p} \\ E_{c}=mgh \\ E_{c}=0.05x9.81x0.95 \\ \boxed {E_{c}=0.465975J}$