A factory worker pushes a 32.0 kg crate a distance of 4.9 m along a level floor at constant velocity by pushing downward at an angle of 32 ∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

What magnitude of force must the worker apply to move the crate at constant velocity?

How much work is done on the crate by this force when the crate is pushed a distance of 4.9 m ?

How much work is done on the crate by friction during this displacement?

What is the total work done on the crate?

To answer these questions, we need to apply Newton's laws and the concept of work.

1. Magnitude of force required to move the crate:
First, we need to find the normal force acting on the crate. Since the crate is on a level floor, the normal force is equal to the weight of the crate, which is given by:

Normal Force = mass × acceleration due to gravity
Normal Force = 32.0 kg × 9.8 m/s^2
Normal Force = 313.6 N

The force required to overcome kinetic friction is given by:

Force of Friction = coefficient of friction × Normal Force
Force of Friction = 0.25 × 313.6 N
Force of Friction = 78.4 N

Now, the horizontal component of the worker's applied force must balance the force of friction to keep the crate at constant velocity. Since the crate is pushed downward at an angle of 32 degrees below the horizontal, we need to find the horizontal component of the worker's force using trigonometry:

Horizontal Component of Force = applied force × cos(angle)
Horizontal Component of Force = F × cos(32°)

Since the crate is moving at constant velocity, the horizontal component of the worker's force is equal to the force of friction:

F × cos(32°) = 78.4 N

We can rearrange this equation to solve for F:

F = 78.4 N / cos(32°)

Now plug in the values and calculate:

F ≈ 92.88 N (rounded to two decimal places)

Therefore, the magnitude of force the worker must apply to move the crate at constant velocity is approximately 92.88 N.

2. Work done on the crate by the worker's force:
The work done by a force is given by the equation:

Work = force × distance × cos(angle)

Since the worker's applied force and the displacement are in the same direction (horizontal), the angle between them is 0 degrees. Therefore, the cos(0°) is equal to 1. Thus, the equation simplifies to:

Work = force × distance

Plugging in the values:

Work = 92.88 N * 4.9 m

Now calculate:

Work ≈ 454.11 J (rounded to two decimal places)

Therefore, the work done on the crate by the worker's force is approximately 454.11 Joules.

3. Work done on the crate by friction:
The work done by friction is given by the equation:

Work = force of friction × distance × cos(angle)

Since the force of friction and displacement are in opposite directions (friction opposes motion), the angle between them is 180 degrees. Therefore, the cos(180°) is equal to -1. Thus, the equation becomes:

Work = force of friction × distance × -1

Plugging in the values:

Work = 78.4 N * 4.9 m * -1

Now calculate:

Work ≈ -383.36 J (rounded to two decimal places)

Note: The negative sign indicates that the work done by friction is in the opposite direction to the displacement.

Therefore, the work done on the crate by friction during this displacement is approximately -383.36 Joules.

4. Total work done on the crate:
The total work done on an object is the sum of the work done by all forces acting on it.

Total Work = Work done by the worker's force + Work done by friction

Plugging in the values:

Total Work = 454.11 J + (-383.36 J)

Now calculate:

Total Work ≈ 70.75 J (rounded to two decimal places)

Therefore, the total work done on the crate is approximately 70.75 Joules.