13. A lever does 5.0 J of work on a 0.10-kg ball bearing in a pinball machine. The ball's resulting speed just after leaving the lever is -
.5 m/s
10 m/s
1 m/s
5 m/s
CLEAR ALL

The resulting speed of the ball is;

10 m/s

Explanation:

The given parameters are;

The amount of work done by the lever = 5.0 J

The mass of the ball on which the work is done, m = 0.10 kg

By the principle of conservation of energy, we have;

The work done by the lever = The kinetic energy. K.E., gained by the ball

The kinetic energy, K.E., is given by K.E. = 1/2·m·v²

Where, for the ball, we have;

m = The mass of the ball

v = The resulting speed of the ball

Therefore, by substituting the known values, we have;

The work done by the lever = 5.0 J = The kinetic energy. K.E., gained by the ball =  1/2 × m × v² =  1/2 × 0.10 kg × v²

∴ 5.0 J = 1/2 × 0.10 kg × v²

v² = 5.0 J/(1/2 × 0.10 kg) = 100 m²/s²

v = √(100 m²/s²) = 10 m/s

The resulting speed of the ball, v = 10 m/s.

small = small current large mag large current

the maxium height is 34m/s