On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 ms⁻¹. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40 kg. Ans. (45 J, 2400 J)


Given:  

Height of the slope = h =  6 m
Speed of the cyclist = v = 1.5 m s⁻¹
Mass of the cyclist = m =  40 kg
gravitational acceleration = g = 10 m s⁻²


To Find:

Kinetic Energy of the cyclist = K.E. = ? 
Potential Energy of the cyclist = P.E. = ? 

Solution:   

The formula for k.E. is 

 K.E.  = `\frac {1}{2}`mv²  

by putting values 

K.E. =  `\frac {1}{2}`(40 kg) (1.5 m s⁻¹)²

K.E. =  `\frac {1}{2}`(40 kg) (2.25 m² s⁻²)

K.E.=  45 kg m² s⁻² 

Or

K.E.=  45 J

the Kinetic energy of the cyclist is 45 J



Now for Potential Energy P.E., we have the formula

P.E.  = mgh

by putting values 

P.E.  = 40 Kg ｘ10 m s⁻² ｘ 6 m

P.E.  = 2400 kg m² s⁻²

or

P.E.  = 120 J

The Potential energy of the cyclist is 45 J

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