Questions: An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 4.7 x 10^4 kg, and the force causes its speed to change from 6200 to 5600 m / s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.2 x 10^6 m determine the magnitude of the force. (a) Work = (b) F=

Solution Steps

We need to determine the work done by a force that slows down an asteroid and then find the magnitude of the force given the distance over which it acts.

The work done by the force can be found using the work-energy principle, which states that the work done is equal to the change in kinetic energy.

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

The change in kinetic energy (\(\Delta K\)) is: \[ \Delta K = K_{\text{final}} - K_{\text{initial}} \] \[ \Delta K = \frac{1}{2} m v_{\text{final}}^2 - \frac{1}{2} m v_{\text{initial}}^2 \]

Substituting the given values: \[ m = 4.7 \times 10^4 \, \text{kg} \] \[ v_{\text{initial}} = 6200 \, \text{m/s} \] \[ v_{\text{final}} = 5600 \, \text{m/s} \]

\[ \Delta K = \frac{1}{2} (4.7 \times 10^4) (5600^2) - \frac{1}{2} (4.7 \times 10^4) (6200^2) \]

\[ \Delta K = \frac{1}{2} (4.7 \times 10^4) (5600^2 - 6200^2) \] \[ \Delta K = \frac{1}{2} (4.7 \times 10^4) (31360000 - 38440000) \] \[ \Delta K = \frac{1}{2} (4.7 \times 10^4) (-7080000) \] \[ \Delta K = (4.7 \times 10^4) (-3540000) \] \[ \Delta K = -1.6638 \times 10^{11} \, \text{J} \]

The work done by the force is: \[ W = \Delta K = -1.6638 \times 10^{11} \, \text{J} \]

The work done by the force is also given by: \[ W = F \cdot d \] where \( F \) is the force and \( d \) is the distance.

Rearranging for \( F \): \[ F = \frac{W}{d} \]

Substituting the given distance: \[ d = 1.2 \times 10^6 \, \text{m} \]

\[ F = \frac{-1.6638 \times 10^{11}}{1.2 \times 10^6} \] \[ F = -1.3865 \times 10^5 \, \text{N} \]

The magnitude of the force is: \[ |F| = 1.3865 \times 10^5 \, \text{N} \]

Final Answer