"Rocket Man" has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and accelerates straight upward. At a height of 16 m, his speed is 5.0 m/s. His mass, including the propulsion unit, has the approximately constant value of 143 kg. Find the work done by the force generated by the propulsion unit.

work= final KE + final PE

= 1/2 m v^2+ mgh

20376.6

To find the work done by the force generated by the propulsion unit, we can use the work-energy theorem.

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

Given that Rocket Man starts from rest and reaches a speed of 5.0 m/s, we can find the change in kinetic energy.

The change in kinetic energy (ΔKE) is given by the formula:

ΔKE = 1/2 * m * (vf^2 - vi^2)

where m is the mass of Rocket Man (143 kg), vi is the initial velocity (0 m/s), and vf is the final velocity (5.0 m/s).

Plugging in the values, we get:

ΔKE = 1/2 * 143 kg * (5.0 m/s)^2

ΔKE = 1/2 * 143 kg * 25 m^2/s^2

ΔKE = 1806.25 kg·m^2/s^2

Since work done (W) is equal to the change in kinetic energy (ΔKE), we can conclude that:

W = 1806.25 kg·m^2/s^2

Thus, the work done by the force generated by the propulsion unit is 1806.25 kg·m^2/s^2.

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

The work done (W) is given by the equation:

W = ΔKE

Where ΔKE is the change in kinetic energy.

Since Rocket Man starts from rest, his initial kinetic energy (KEi) is zero. At a height of 16 m, his final speed (vf) is 5.0 m/s. Therefore, his final kinetic energy (KEf) can be calculated using the equation:

KEf = (1/2) * m * v^2

Where m is the mass of Rocket Man and v is the final velocity.

Substituting the given values:

KEf = (1/2) * 143 kg * (5.0 m/s)^2

Simplifying:

KEf = 1,787.5 J

The change in kinetic energy (ΔKE) is then given by:

ΔKE = KEf - KEi

Since KEi = 0, ΔKE = KEf.

Therefore, ΔKE = 1,787.5 J

So, the work done by the force generated by the propulsion unit is 1,787.5 Joules.