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Physics
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
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Best Answer

Answer:

a)

[tex] - 7.04\times10^{11} J[/tex]

b)

[tex] 5.03\times10^{5} N[/tex]

Explanation:

a)

[tex]m[/tex] = mass of the asteroid = 43000 kg

[tex]v_{o}[/tex] = initial speed of asteroid = 7600 m/s

[tex]v[/tex] = final speed of asteroid = 5000 m/s

[tex]W[/tex] = Work done by the force on asteroid

Using work-change in kinetic energy theorem

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

b)

[tex]F[/tex] = magnitude of force on asteroid

[tex]d[/tex] = 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

[tex]W = - F d\- 7.04\times10^{11} = - F (1.4\times10^{6})\F = 5.03\times10^{5} N[/tex]

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