A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 5.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do

1568 J

Explanation:

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

Mass (m) = 80 Kg

Distance (d) = 5 m

Coefficient of kinetic friction (μ) = 0.4

Workdone (Wd) =?

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

Mass (m) = 80 Kg

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

Normal reaction (N) =?

N = mg

N = 80 × 9.8

N = 784 N

Next, we shall determine force of friction. This can be obtained as follow:

Coefficient of kinetic friction (μ) = 0.4

Normal reaction (N) = 784 N

Force of friction (F) =?

F = μN

F = 0.4 × 784

F = 313.6 N

Finally, we shall determine the work done. This can be obtained as follow:

Distance (d) = 5 m

Force of friction (F) = 313.6 N

Workdone (Wd) =?

Wd = F × d

Wd = 313.6 × 5

Wd = 1568 J

Thus, the workdone is 1568 J

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