Questions: 9. When positive work is done on an object, its kinetic energy increases; negative work is done on it, its kinetic energy decreases. 10. How much work is done to stop a 1000-kg car moving at 30 m/s? Be sure to include the correct sign with your answer. 11. If the speed of an object doubles, its kinetic energy increases by a factor. 12. Define what is meant by potential energy.

When positive work is done on an object, its kinetic energy increases; when negative work is done on it, its kinetic energy decreases. \(\boxed{\text{increases, decreases}}\)

The work done to stop a car is equal to the change in kinetic energy. The initial kinetic energy \( KE_i \) of the car is given by: \[ KE_i = \frac{1}{2} m v^2 \] where \( m = 1000 \, \text{kg} \) and \( v = 30 \, \text{m/s} \).

Substituting the values: \[ KE_i = \frac{1}{2} \times 1000 \, \text{kg} \times (30 \, \text{m/s})^2 = 450,000 \, \text{J} \]

Since the car is brought to a stop, the final kinetic energy \( KE_f = 0 \). The work done \( W \) is: \[ W = KE_f - KE_i = 0 - 450,000 \, \text{J} = -450,000 \, \text{J} \]

The negative sign indicates that the work is done against the motion of the car.

The work done to stop the car is \(\boxed{-450,000 \, \text{J}}\).

The kinetic energy \( KE \) of an object is given by: \[ KE = \frac{1}{2} m v^2 \]

If the speed \( v \) doubles, the new speed is \( 2v \). The new kinetic energy \( KE' \) is: \[ KE' = \frac{1}{2} m (2v)^2 = \frac{1}{2} m \times 4v^2 = 4 \times \frac{1}{2} m v^2 = 4 \times KE \]

Thus, the kinetic energy increases by a factor of 4.

If the speed of an object doubles, its kinetic energy increases by a factor of \(\boxed{4}\).