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  

mgh12 m v² + f s

Or

12 m v² = mgh -f s

or

v(2mgh-fs)m


by putting the corresponding values

v = 
(2100kg9.8ms²10m-480N100m)100kg

v = (98000kgm²s²-48000Nm)100kg

v = 50000kgm²s²100kg

v = 100m²s²

v = 10 m s⁻¹ -------------Ans.


************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK    

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.