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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. (45 J, 2400 J)
Solution:
Height of the slope = h = 6 m
Speed of the cyclist = v = 1.5 $ms^{-1}$
Mass of cyclist and the bicycle = m = 40 kg
Kinetic energy = K.E. = ?
Potential energy = P.E. = ?
(i) K.E. = $\frac{1}{2}m v ^{2}$
K.E = $\frac{1}{2}\times 40\times (1.5)^{2}$
K.E = $\frac{1}{2}\times 40\times 2.25$ = 45 J
(ii) P.E = mgh
P.E = $40 \times 10 \times 6$ = 2400 J