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/s². 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/s²

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 v² = 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 v² = 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/s², 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_t² – v_o²) = 1/2 m (v_t² – 0) = 1/2 m v_t²

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_t² – 400 m

1/2 m v_t² = 400 m

1/2 v_t² = 400

v_t² = 2(400)

v_t = √(400)(2)

v_t = 20√2 m/s