A net force of 25 N is applied for 5.7 s to a 12-kg box initially at rest. What is the speed of the box at the end

Question

Topic: Physics Posted 9 months ago

A net force of 25 N is applied for 5.7 s to a 12-kg box initially at rest. What is the speed of the box at the end of the 5.7-s interval?

A) 1.8 m/s
B) 12 m/s
C) 3.0 m/s
D) 7.5 m/s
E) 30 m/s

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Mar 18, 2024

Koala Ai Support

To find the speed of the box at the end of the 5.7-s interval, we first need to calculate the acceleration of the box using Newton's second law:

F = ma
25 N = 12 kg * a
a = 25 N / 12 kg
a = 2.08 m/s^2

Next, we can use the equation for acceleration to find the final velocity of the box:

vf = vi + at
vf = 0 m/s + 2.08 m/s^2 * 5.7 s
vf ≈ 11.856 m/s

Therefore, the speed of the box at the end of the 5.7-s interval is approximately 11.856 m/s. Since this falls closest to 12 m/s, the answer is:

B) 12 m/s

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