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?

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.
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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