# Conservation of mechanical energy on curve surface – problems and solutions

1. If ball’s speed at point A is 6 m/s, ball’s speed at point B is √92 m/s, and acceleration due to gravity is g = 10 m/s^{2}. What is the height of point B (h)?

__Known :__

Speed of ball at point A (v_{A}) = 6 m/s

Speed of ball at point B (v_{B}) = √92 m/s

Acceleration due to gravity (g) = 10 m/s^{2}

The height of A (h_{A}) = 5.6 meters

The height of B (h_{B}) = h

__Wanted :__ The height of point B (h)

__Solution :__

**The initial mechanical energy ****= the gravitational potential energy**

At point A, ball’s speed = 0, so the kinetic energy of the ball = 0. KE = 1/2 m v^{2 }= 1/2 m (0) = 0.

Ball at the height of 5.6 meters so the ball has the gravitational potential energy. The gravitational potential energy : PE = m g h = m (10)(5.6) = 56 m

*The initial mechanical energy = the gravitational potential energy + kinetic energy = **56 m + 0 = 56 m *

**The final mechanical energy = the gravitational potential energy + kinetic energy**

At point B, the height of ball is h. The gravitational potential energy : PE = m g h = m (10) h = 10 m h

The kinetic energy of the ball : KE = 1/2 m v^{2 }= 1/2 m (√92)^{2 }= 1/2 m (92) = 46 m

*The final mechanical energy = the gravitational potential energy **+ the kinetic energy = 10 m h + 46 m = m (10 h + 46)*

**The principle of conservation of mechanical energy :**

*The initial mechanical energy = the final mechanical energy*

56 m = m (10 h + 46)

56 = 10 h + 46

56 – 46 = 10 h

10 = 10 h

h = 10/10

h = 1 meter

2.

If the initial velocity = 0, the acceleration due to gravity = 10 m/s^{2}, then what is the speed at the height of B.

__Known :__

The initial speed (v_{o}) = 0

The initial height (h_{o}) = 50 m – 10 m = 40 m

The final height (h_{t}) = 0

Acceleration due to gravity (g) = 10 m.s^{-2}

__Wanted :__ The final speed (v_{t})

__Solution :__

The change of the kinetic energy :

ΔKE = 1/2 m (v_{t}^{2} – v_{o}^{2}) = 1/2 m (v_{t}^{2} – ) = 1/2 m v_{t}^{2}

The change of the potential energy :

ΔPE = m g (h_{t} – h_{o}) = m (10)(0-40) = m (10)(-40) = – 400 m

Calculate the final speed (v_{t}) using the equation of the principle of conservation of mechanical energy :

0 = ΔKE + ΔPE

0 = 1/2 m v_{t}^{2} – 400 m

1/2 m v_{t}^{2} = 400 m

1/2 v_{t}^{2} = 400

v_{t}^{2} = 2(400)

v_{t} = √(400)(2)

v_{t} = 20√2 m/s