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A worker pushed a 25kg crate over a distance 6 meters horizontally at a constant velocity. the coefficient of kinetic friction is 0.300.

what magnitude of force must the worker apply?

1 Answer

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Step-by-step explanation:

To maintain a constant velocity, the worker must apply a force that is equal in magnitude and opposite in direction to the force of kinetic friction acting on the crate. The force of kinetic friction can be found using the formula:

f_k = μ_k * N

where f_k is the force of kinetic friction, μ_k is the coefficient of kinetic friction, and N is the normal force, which is equal in magnitude to the weight of the crate:

N = m * g

where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.81 m/s²).

Substituting the given values, we get:

N = m * g = 25 kg * 9.81 m/s² = 245.25 N

f_k = μ_k * N = 0.300 * 245.25 N = 73.575 N

Therefore, the worker must apply a force of 73.575 N in the horizontal direction to maintain a constant velocity.

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