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 3.7× 104 kg, and the force causes its speed to change from 7500 to 4500m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.7× 106 m determine the magnitude of the force.

Answer 1

Answer:

666000000000 J

391764.71 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

Change in energy

$\Delta KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow \Delta KE=\frac{1}{2}3.7\times 10^4(4500^2-7500^2)\\\Rightarrow \Delta KE=-666000000000\ J$

Work done be the force is 666000000000 J

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{-666000000000}{1.7\times 10^6}\\\Rightarrow F=-391764.71\ N$

The magnitude of force is 391764.71 N