Richa weighing 40 kgf leaves point P on her skateboard and reaches point Q on the ground with velocity 10 m s⁻¹. Calculate:
(a) the kinetic energy of Richa at point Q
(b) the vertical height of point P above the ground. (Use g as 10 m s⁻² and neglect friction.)
(c) the kinetic energy of Richa at point R. (While moving from Q to R, she loses 500 J of energy against friction.)
(a) Given, speed at Q (v) = 10 m s⁻¹
Kinetic energy at Q = \frac{1}{2} \times40 \times10^2 = 2000 J
(b) If the height of P is H, then potential energy at P = 40 × 10 × H = 400H.
As friction is negligible, total mechanical energy is conserved.
400H = 2000
H = 5 m
(c) Height of point R = 3 m and energy lost from Q to R = 500 J.
Potential energy at R = 40 × 10 × 3 = 1200 J.
Kinetic energy at R = 2000 − 1200 − 500 = 300 J.
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