Question

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 3.6× 10⁴ kg, and the force causes its speed to change from 7100 to 5200 m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 2.4x 10⁶ m, determine the magnitude of the force?

Answer 1

The work done by the force on the asteroid is -4.2099 × 10ⁱ² J, and the magnitude of the force is 17541.25 N acting in the opposite direction of the displacement.

Step-by-step explanation:

To calculate the work done by the force on the asteroid:

Work done (W) = Change in kinetic energy = ½ m (v²₂ - v²₁)

Where m is the mass of the asteroid, v₂ is the final speed, and v₁ is the initial speed.

Substituting the given values: m = 3.6 × 10⁴ kg, v₁ = 7100 m/s, and v₂ = 5200 m/s:

W = ½ × (3.6 × 10⁴ kg) × (5200² - 7100²) m²/s²

W = ½ × (3.6 × 10⁴ kg) × (27040000 - 50410000) m²/s²

W = ½ × (3.6 × 10⁴ kg) × (-23370000) m²/s²

W = -4.2099 × 10ⁱ² J

To find the magnitude of the force (F), we use the work-energy principle:

Work done (W) = Force (F) × Displacement (s)

W = F × s

Where s is the displacement over which the force acts.

Substituting W = -4.2099 × 10ⁱ² J and s = 2.4× 10⁶ m:

F = W / s

F = (-4.2099 × 10ⁱ² J) / (2.4 × 10⁶ m)

F = -17541.25 N

The negative sign indicates that the force is in the opposite direction of the displacement, which aligns with the fact that the force slows down the asteroid.