A 1380 kg car starts from rest at the top of a 28.0 m long hill inclined at 11.00 degrees. Ignoring friction, how fast is it going when it reaches the bottom of the hill?​

A 1380 kg car starts from rest at the top of a 280 m long hill inclined at 1100 degrees Ignoring friction how fast is it going when it reaches the bottom of the class=

Respuesta :

Answer:

10.2 m/s

Explanation:

Using conservation of energy:

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 9.8 m/s² × 28.0 m × sin 11.0°)

v = 10.2 m/s

The velocity when car reaches the bottom of the hill is 10.2 m/s.

What is mechanical energy?

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

M.E = ½ mv² + mgh

where g is the acceleration due to gravity, v is the velocity, m is the mass and h is the height of the object.

Given is a 1380 kg car starts from rest at the top of a 28.0 m long hill inclined at 11.00 degrees.

Using conservation of energy principle, we have

PE = KE

mgh = ½ mv²

v = √(2gh)

Substitute the values, we get

v = √(2 × 9.8 m/s² × 28.0 m × sin 11.0°)

v = 10.2 m/s

Thus, the velocity when car reaches the bottom of the hill is 10.2 m/s.

Learn more about mechanical energy.

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