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 4.3 × 104 kg, and the force causes its speed to change from 7600 to 5000 m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.4 × 106 m determine the magnitude of the force.
I have part A, just need part B
A= -7.048E11 J

Answer 1

Answer:

a)

$- 7.04\times10^{11} J$

b)

$5.03\times10^{5} N$

Explanation:

a)

$m$ = mass of the asteroid = 43000 kg

$v_{o}$ = initial speed of asteroid = 7600 m/s

$v$ = final speed of asteroid = 5000 m/s

$W$ = Work done by the force on asteroid

Using work-change in kinetic energy theorem

$W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J$

b)

$F$ = magnitude of force on asteroid

$d$ = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

$W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N$