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]
(a) The work done by force be "-7.04 × 10¹¹ J".
(b) The magnitude of force be "5.03 × 10⁵ N".
Force and Displacement
According to the question,
Asteroid's mass, m = 43000 kg
Asteroid's initial speed, [tex]v_0[/tex] = 7600 m/s
Asteroid's final speed, v = 5000 m/s
Distance travelled by asteroids, d = 1.4 × 10⁶ m
Now,
(a) By using work change in Kinetic energy theorem, we get
W = [tex]\frac{1}{2}[/tex]m(v² - v₀²)
By substituting the values,
= [tex]\frac{1}{2}[/tex] × 43000 (5000)² - (7600)²
= -7.04 × 10¹¹ J
(b) We know that,
W = -Fd
-7.04 × 10¹¹ = -F(1.4 × 10⁶)
F = 5.03 × 10⁵ N
Thus the above answer is correct.
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