A 1000 kg automobile at the top of an inclined plane 10 m high and 100 m long is released and rolls down the hill. What is its speed at the bottom of inclined if the average retarding force due to friction is 480 N? (Ans: 10 m s⁻¹)
Given:
Mass of Automobile = m = 40 g = 0.04 kg
Height of the inclined plane = h = 10 m
Length of the plane = S = 100 m
Retarding force = f = 480 N
To Find:
Velocity of Automobile = v = ?
Solution:
By using Law of conservation of Energy
Loss of P.E. = Gain in K.E. + work done against friction
mgh = 12 m v² + f s
Or
12 m v² = mgh -f s
or
v = √(2mgh-fs)m
by putting the corresponding values
v = √(98000kgm²s⁻²-48000Nm)100kg
v = √50000kgm²s⁻²100kg
v = √100m²s⁻²
v = 10 m s⁻¹ -------------Ans.
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