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.)

(a) Given,

Speed at Q (v) = 10 ms^-1

Then,

Kinetic energy at Q = $\dfrac{1}{2}$ mv² = $\dfrac{1}{2}$ × 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 = $\dfrac{2000}{400}$ = 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