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rocket man has a propulsion unit strapped to his back. he starts from rest on the ground, fires the unit, and ispropelled straight upward. at a height of 16 m, his speed is 5.0 m/s. his mass, including the propulsion unit, hathe approximately constant value of 136 kg. find the work done by the force generated by the propulsion unit.

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Answer:

the work done by the force generated by the propulsion unit is 1700 kg * m^2/s^2.

Explanation:

To find the work done by the force generated by the propulsion unit, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

Given:

Initial speed (u) = 0 m/s (starting from rest)

Final speed (v) = 5.0 m/s

Height (h) = 16 m

Mass (m) = 136 kg

First, let's find the change in kinetic energy:

Change in kinetic energy (ΔKE) = (1/2) * m * (v^2 - u^2)

Substituting the values:

ΔKE = (1/2) * 136 kg * (5.0 m/s)^2 - 0 m/s)^2)

ΔKE = (1/2) * 136 kg * (25 m^2/s^2 - 0 m^2/s^2)

ΔKE = (1/2) * 136 kg * 25 m^2/s^2

ΔKE = 1700 kg * m^2/s^2

Now, since work (W) is equal to the change in kinetic energy, we have:

Work done by the force generated by the propulsion unit (W) = ΔKE

W = 1700 kg * m^2/s^2

Therefore, the work done by the force generated by the propulsion unit is 1700 kg * m^2/s^2.

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