Question

Richa weighing 40 kgf leaves point P on her skateboard and reaches point Q on the ground with velocity 10 ms⁻¹. Calculate:

1. the kinetic energy of Richa at point Q.
2. the vertical height of point P above the ground. (Use g as 10 m/s² and neglect friction)
3. the kinetic energy of Richa at point R. (While moving from Q to R, she loses 500 J of energy against friction.)

Answer 1

a. Kinetic energy = `1/2 mv^2`

= `1/2 xx 40 xx 10 xx 10`

= 2000 J

b. By the law of conservation of energy:

KE at Q = PE at P

2000 = m × g × h

h = `2000/(40 xx 10)`

= 5 m

c. Now,

KE at R = 2000 − PE at R − Energy lost due to friction

= 2000 − (40 × 10 × 3) − 500

= 300 J