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 = `\frac {1}{2}` m `\v²` + f s

Or

`\frac {1}{2}` m `\v²` = mgh -f s

or

v = `\sqrt frac {(2mgh - fs)}{m}`


by putting the corresponding values

v = `\sqrt frac {(2ｘ100 kg ｘ 9.8 m s⁻² ｘ10 m - 480 N ｘ 100 m)}{100 kg}`

v = `\sqrt frac {(98000 kg m² s⁻² - 48000 N m)}{100 kg}`

v = `\sqrt frac {50000 kg m² s⁻²}{100 kg}`

v = `\sqrt {100 m² s⁻²}`

v = 10 m s⁻¹ -------------Ans.


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