Calculating Work Done on a 50kg Crate by 100N Force

[Drey0287]

A 50-kg crate is pulled 40 m along a horizontal floor by a constant 100 N force exerted by a person which acts at a 37 degree angle above the horizontal. Determine the work done on the crate.

I know the equation for work is

work = F * delta X (triangleX) OR work = m*acceleration*deltaX

But where do I use the 37 degree angle?

 

[ascky]

You have to take into account the direction of the force
W=Fxcos(theta)

 

[dextercioby]

A 50-kg crate is pulled 40 m along a horizontal floor by a constant 100 N force exerted by a person which acts at a 37 degree angle above the horizontal. Determine the work done on the crate.

I know the equation for work is

work = F * delta X (triangleX) OR work = m*acceleration*deltaX

But where do I use the 37 degree angle?

If the force would have acted parallel to the horizontal,u could have stated that the work is simply the product between the force's magnitude (100 N) and the distance (40m).That would not be the case,though,because not all of that force is acting on the crate.Only the horizontal component does.Then u should check whether the vertical component annulates the pression force on the ground as to lift the crate.Simple manipulation of numbers shows you that it doesn't.
Hopefully u're clear now.

 

[Raza]

Question:
A 50-kg crate is pulled 40 m along a horizontal floor by a constant 100 N force exerted by a person which acts at a 37 degree angle above the horizontal. Determine the work done on the crate.

Solution:
Work=(Force)(Displacment)(COS degrees)

$$W=fdcos$$

$$=(100)(40)(Cos 37)$$

$$=(100)(40)(0.80)$$

$$=3200 J$$

$$=3.2 kJ$$

I might be wrong.