The factory worker pushes a 30.0 kg crate a distance of 3.3 mm along a level floor at a constant velocity by pushing downward at an angle of 30 ∘∘ below the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. What magnitude of force must the worker apply to move the crate at constant velocity?

The magnitude of force is 99.2 N.

Given:

m=30.0kg

To solve the problem, we have to write the equation of the forces along two perpendicular directions.

OR

................(i)

where,

N is the normal reaction of the floor on the crate

F is the force applied by the worker

is the angle at which the force is applied

mg is the weight of the crate, with

m = 30.0 kg (mass of the crate)-given

(acceleration of gravity)

where,

is the force of friction, where

is the coefficient of friction-given

a is the acceleration; since the crate is moving at constant velocity, the acceleration is zero,

∴...............(ii)

on equating (i) and (ii)

The magnitude of force is 99.2 N.

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99.2 N

Explanation:

To solve the problem, we have to write the equation of the forces along two perpendicular directions.

Along the vertical direction:

(1)

where

N is the normal reaction of the floor on the crate

F is the force applied by the worker

is the angle at which the force is applied

mg is the weight of the crate, with

m = 30.0 kg (mass of the crate)

(acceleration of gravity)

Along the horizontal direction:

where

is the force of friction, where

is the coefficient of friction

and

a is the acceleration; since the crate is moving at constant velocity, the acceleration is zero, so

And the equation becomes

(2)

From eq.(1) we get:

And substituting into (2) we get

6/12=1/2

step-by-step explanation:

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