Richa weighing 40 kgf leaves point P on her skateboard and reaches point Q on the ground with velocity 10 ms-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/s2 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.)
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(a) Given,
Speed at Q (v) = 10 ms-1
Then,
Kinetic energy at Q = $\frac{1}{2}$ mv2 = $\frac{1}{2}$ × 40 × 102 = 2000 J
(b) If the height of P is H, then :
Potential energy at P = mgH = 40 x 10 x H = 400H
As, friction is negligible so total mechanical energy is conserved.
From law of conservation of total mechanical energy,
Potential energy at P = Kinetic energy at Q
⇒ 400H = 2000 J
⇒ H = $\frac{2000}{400}$ = 5 m
(c) Given,
Height of point R (h) = 3 m
Energy lost from Q to R through friction = 500 J
Now
Potential energy at R = mgh = 40 × 10 × 3 = 1200 J
So, from law of conservation of energy
Kinetic energy at R + Potential energy at R + Energy lost from Q to R through friction = Potential energy at P
⇒ Kinetic energy at R = Potential energy at P - Potential energy at R - Energy lost from Q to R through friction
⇒ Kinetic energy at R = 2000 - 1200 - 500 = 300 J