A 151 kg crate is pulled along a level surface by an engine. The coefficient of kinetic friction between the crate and the surface is 0.399. How much power must the engine deliver to move the crate at a constant speed of 6.04 m/s

3566.26 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) of crate = 151 kg

Coefficient of kinetic friction (μ) = 0.399

Velocity (v) = 6.04 m/s

Power (P) =?

Next, we shall determine the normal reaction. This can be obtained as follow:

Mass (m) of crate = 151 kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) = mg

Normal reaction (R) = 151 × 9.8

Normal reaction (R) = 1479.8 N

Next, we shall determine the force applied to move the crate. This can be obtained as follow:

Coefficient of kinetic friction (μ) = 0.399

Normal reaction (R) = 1479.8 N

Force (F) =?

F = μR

F = 0.399 × 1479.8

F = 590.44 N

Finally, we shall determine the power used to move the crate as follow:

Velocity (v) = 6.04 m/s

Force (F) applied = 590.44 N

Power (P) =?

P = F × v

P = 590.44 × 6.04

P = 3566.26 Watts

Therefore, the power used to move the crate is 3566.26 Watts.

we need answer choices

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up