Question

A crate with a mass of 60 kg glides through a space station with a speed of 2 m/s. An astronaut speeds it up by pushing on it from behind with a force of 180 N, continually pushing with this force through a distance of 7 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 160 N, backing up through a distance of 6 m. After these two maneuvers, what is the speed of the crate?

Answer 1

Answer:

$v_f = 3.74 m/s$

Explanation:

As per work energy theorem we know that

work done by all forces = change in kinetic energy

now we have

$W_1 + W_2 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

$W_1 = 180 \times 7 = 1260 J$

$W_2 = 160 \times 6 = - 960 J$

now we have

$1260 - 960 = \frac{1}{2}(60)(v_f^2 - 2^2)$

$300 = 30(v_f^2 - 4)$

$14 = v_f^2$

$v_f = 3.74 m/s$

Answer 2

Work done is the change in the kinetic energy of an object. After the two maneuvers, the speed of the crate will be √14 m/s.

What is work done?

Work done is the change in the kinetic energy of an object.

$W=\dfrac{1}{2}m(v_1^2-v_o^2)$

We know that the change in the kinetic energy of an object is the change in the velocity, also, the work done is the product of force and displacement. therefore, the work done can be written,

$\dfrac{1}{2} \times 60 \times (v_1^2-v_o^2) = (F_1\cdot ds) - (F_2 \cdot ds)\\\\\dfrac{1}{2} \times 60 \times (v_1^2-2^2) = (180 \times 7) - (160 \times 6)\\\\30 \times (v_1^2-4) = 300\\\\v_1^2 =10+4\\\\v_1 = \sqrt{14}$

Hence, After the two maneuvers, the speed of the crate will be √14 m/s.

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