A 1.60 102-kg crate is being pushed across a horizontal floor by a force that makes an angle of 30.0° below the horizontal. The coefficient of kinetic friction is 0.200. What should be the magnitude of , so that the net work done by it and the kinetic frictional force is zero?

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To find the magnitude of the force, we need to set up an equation that considers the forces acting on the crate and the work done by them.

Let's analyze the forces acting on the crate first:

1. The force pushing the crate: We'll call it F. It's acting at an angle of 30.0° below the horizontal.

2. The weight of the crate: We'll call it mg, where m is the mass of the crate (102 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).

3. The frictional force: We'll call it f. Since the crate is in motion, we are dealing with kinetic friction. The frictional force can be calculated using the coefficient of kinetic friction (µk) and the normal force.

The normal force can be calculated as the vertical component of the weight, which is mg * cos(30°).

Now, let's set up the equation:

The net work done is zero, which means the work done by the pushing force F should be equal and opposite to the work done by the frictional force f.

Work done by F: F * d * cos(180° - 30°), where d is the distance the crate is pushed.

Work done by f: f * d * cos(180°), where cos(180°) = -1, as the work done by friction acts in the opposite direction.

Setting up this equation, we get:

F * d * cos(180° - 30°) = -f * d * cos(180°)

Using the definition of the coefficient of kinetic friction, f = µk * Normal force, we can substitute it in the equation:

F * d * cos(180° - 30°) = -µk * Normal force * d * cos(180°)

Substituting the values:

F * d * cos(150°) = -0.200 * (mg * cos(30°)) * d * cos(180°)

Simplifying the equation:

F * cos(150°) = -0.200 * mg * cos(30°)

Finally, solving for the magnitude of F, we get:

F = (-0.200 * mg * cos(30°)) / cos(150°)

Plugging in the values:

F ≈ (-0.200 * (102 kg * 9.8 m/s²) * cos(30°)) / cos(150°)

F ≈ - 156.75 N

The magnitude of the force should be approximately 156.75 N (pointing in the opposite direction of the pushing force) to make the net work done by it and the kinetic frictional force zero.