Work: Kinetic Friction + Applied Resultant Force

[HumKinStudent]

Homework Statement

The coefficient of kinetic friction between a 20kg box and the floor is 0.40 -> How much work does a pulling force applied 37 degrees above the horizontal do on the box in pulling it 8.0m across the floor at constant speed

Homework Equations

F(kinetic)= F(normal) x U(kinetic)

W=Fxd

Fx=Fcos(theta)

The Attempt at a Solution

I solved for kinetic friction, I'm just not 100% sure where to go next, a hint would be more helpful in the long run than the answer to the entire question... I was thinking of assuming that force in the x direction should be equal to F(kinetic) and then just multiplying f(kinetic) by the distance 8 metres... Nonetheless I feel like there is more to it...

 

[kuruman]

I'm just not 100% sure where to go next, a hint would be more helpful in the long run than the answer to the entire question... I was thinking of assuming that force in the x direction should be equal to F(kinetic) and then just multiplying f(kinetic) by the distance 8 metres... Nonetheless I feel like there is more to it...

There cannot be a single force in the x-direction if the box is moving at constant speed. Do you see why?

 

[HumKinStudent]

There cannot be a single force in the x-direction if the box is moving at constant speed. Do you see why?

if the box is moving at constant speed in the x direction then the force MUST be equal to the kinetic friction, correct?

Should I go back, make my sum of Forces in the y = 0 and recalculate the normal force by factoring in the y component of the force, instead of assuming the normal force is the opposite of the box's weight x gravity?

This will change my kinetic force of friction...

 

[kuruman]

if the box is moving at constant speed in the x direction then the force MUST be equal to the kinetic friction, correct?

It is correct to say that, because the speed is constant, the component of the pulling force in the horizontal direction and the force of kinetic friction add to zero.

Should I go back, make my sum of Forces in the y = 0 and recalculate the normal force by factoring in the y component of the force, instead of assuming the normal force is the opposite of the box's weight x gravity?

This will change my kinetic force of friction...

Yes you should and consider the component of the pulling force perpendicular to the incline.

 

[HumKinStudent]

It is correct to say that, because the speed is constant, the component of the pulling force in the horizontal direction and the force of kinetic friction add to zero.



Yes you should and consider the component of the pulling force perpendicular to the incline.

there is no incline in this question, it is on a flat surface, nonetheless, I should still break down the resultant force...

 

[kuruman]

Yes, there is no incline, I got confused. You should break down the puling force into horizontal and vertical components. Since the acceleration is zero, the sum of all the horizontal components must be zero and the sum of all the vertical components must be zero.