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A 213 kg crate slides 7.6 m until coming to a stop after being pushed by a large man.

The effective coefficient of kinetic friction between the crate and the surface is 0.37.
Calculate the work done by friction.
A)1,133.16 J
B)772.34 J
C)2,087.4 J
D)5,869.78 J

User Keith Morris
by
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2 Answers

23 votes
23 votes

Final answer:

The work done by friction is calculated using the kinetic friction coefficient, mass, acceleration due to gravity, and distance. The correct work calculation reveals that none of the options provided matches the correct answer of -6071.7828 J.

Step-by-step explanation:

To calculate the work done by friction, we use the formula Work = force × distance × cos(θ), where θ is the angle between the force and direction of motion. Since friction acts in the opposite direction to the motion, θ is 180°, and cos(180°) is -1. The force of friction (f) is the product of the coefficient of kinetic friction (μ) and the normal force (N), which is equal to the weight of the crate in this case (mass × acceleration due to gravity). The formula for frictional force is thus: f = μ × m × g. Therefore, the work done by friction (Wfr) can be given as Wfr = μ × m × g × d × (-1).

Plugging the given values into this formula:

  • Mass (m) of the crate = 213 kg
  • Distance (d) slid = 7.6 m
  • Coefficient of kinetic friction (μ) = 0.37
  • Acceleration due to gravity (g) = 9.81 m/s2

Then, Wfr = 0.37 × 213 kg × 9.81 m/s2 × 7.6 m × (-1) = -6071.7828 J. The negative sign indicates that the work is done against the movement of the crate. Since options A, B, C, and D do not match the calculated value, none of the provided options is correct. The correct answer should have been -6071.7828 J, indicating energy loss due to friction.

User Bravokiloecho
by
3.1k points
29 votes
29 votes

Answer:D

Step-by-step explanation:

User Eugenio De Hoyos
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3.0k points