Afactory worker pushes a 30.0kg crate a distance of 4.4 m along a level floor at constant velocity by pushing downward at an angle of 31 below the horizontal. the coefficient of kinetic friction between the crate and floor is 0.24.

1-what magnitude of force must the worker apply to move the crate at constant velocity?
2-how much work is done on the crate by this force when the crate is pushed a distance of 4.4m?
3-how much work is done on the crate by friction during this displacement?
4-how much work is done by the normal force?
5-how much work is done by gravity?
6-what is the total work done on the crate?

1)  F = 94.58 N , 2) W1 = 360.4 J , 3)  W2 = -360.4 J , 4) W3 =0

Explanation:

1- To find the value of the force, let's use Newton's second law, where in the x-axis we have the applied force and the friction force that opposes the movement and the axis and we have the normal and the weight, look at the attached to see A free body diagram.

We see that the applied force (F) must be decomposed

     Cos θ = Fx / F

     Fx = f cos θ

     sinθ = Fy / F

     Fy = F sin θ

As the box moves at constant speed the acceleration is zero

X axis

      Fx-fr 0 = 0

      Fx = fr

      fr = μ N

      F cos T = μ N

      F cos T = μ N

Axis y

      N -W- Fy = 0

      N- F sin θ = mg

Let's write the equation system and solve it

      F cos θ = μ N

      N - F sin θ = mg                

      N = mg + F sinθ         (1)

      F cos θ = μ (mg + F sin θ)

      F (cos θ - μ Sin θ) = μ mg

      F (cos 30 - 0.24 sin 30) = 0.24 30.0 9.8

      F = 70.56 / 0.746

      F = 94.58 N

This is the value of the force applied

2- The work is defined as the scalar product of the force by the distance traveled, in this case as we have the components of the force the only one that performs work is the Enel X-axis component, because with the other the angle is 90º and the cosine of 90 is zero

      W1 = Fx X

      W1 = 94.58 cos 30 4.4

      W1 = 360.4 J

3- Let's calculate the value of the friction force

      fr = μ N

The value of normal can be found in equation 1

      N = mg + F sin θ

      N = 30.0 9.8 + 94.58 sin 30

      N = 341.3 N

      fr = 0.24 341.3

      fr = 81.9 N

Now we can calculate the work, but let's look at the angle between the displacement and the friction force that is 180º so the cosine of 180 = -1

     W2 = - fr X

     W2 = - 81.9 4.4

     W2 = -360.4 J

4- The normal one has a direction perpendicular to the displacement, so its angle is 90º and the cos 90 = 0, which implies that the work is zero

        W3 =0

5- To calculate total work we just have to add the work of each force

      W all = W1 + W2 + WN + Ww

      W all = 360.4 + (- 360.4) + 0+ 0

      W all = 0

show the diagram then i will answer

the results is a decrease in air temperature with an increase of altitude.

explanation: