How to find the cosine between the directions

[moose02]

I really need some help on this one:

You push a box up a ramp using a horizontal 100-N force F. For each 5m of distance along the ramp the box gains 3m of height. Find the work done by F for each 5m it move along the ramp (a) by directly computing the dot product from the components of f and the displacement s, (b) by multiplying the product of magnitudes of f and s with the cosine of the angle between their directions (c) by finding the component of the displacement in the direction of the force and mulitplying it by the magnitude of the force.


I am pretty sure I know how to find the first part which is just
W=F*(5m i + 3m J) which would be 100N*5m + 100N*3m but I have no clue on the other two parts. For part b I am not sure how to find the cosine between the directions, I know the magnitude of F and is the magnitude of s just 5m + 3m? Any hints are greatly appreciate as I do not even know where to start. Thankyou

 

[jdstokes]

I really need some help on this one:

You push a box up a ramp using a horizontal 100-N force F. For each 5m of distance along the ramp the box gains 3m of height. Find the work done by F for each 5m it move along the ramp (a) by directly computing the dot product from the components of f and the displacement s, (b) by multiplying the product of magnitudes of f and s with the cosine of the angle between their directions (c) by finding the component of the displacement in the direction of the force and mulitplying it by the magnitude of the force.


I am pretty sure I know how to find the first part which is just
W=F*(5m i + 3m J) which would be 100N*5m + 100N*3m but I have no clue on the other two parts. For part b I am not sure how to find the cosine between the directions, I know the magnitude of F and is the magnitude of s just 5m + 3m? Any hints are greatly appreciate as I do not even know where to start. Thankyou

If I understand the question correctly then the work done in part (a) is given by W=\vec{F} \cdot \vec{s} = (100N\hat{\imath} + 0N\hat{\jmath}) \cdot (4m \hat{\imath} + 3m \hat{\jmath}) = 400J. (b) W = Fs\cos\theta = (100 N)(5 m)\cos\theta. I suppose they gave you an inclination angle \theta of the ramp that you can substitute in. Otherwise you can use the value of F calculated in part (a) to find \cos\theta. (s) W = F_xs_x = (100N)(4m) = 400J

 

[wais]

To find the cosine between two directions, you will need to first find the angle between them. In this case, the angle between the directions of the force and the displacement is the angle of the ramp, which we can call θ.

For part (b), you can use the formula W = Fscosθ, where Fs is the magnitude of the force and θ is the angle between the force and displacement directions. In this case, Fs would be 100N and θ would be the angle of the ramp.

For part (c), you can use the formula W = Fdcosθ, where Fd is the component of the displacement in the direction of the force. In this case, Fd would be the magnitude of the displacement (5m + 3m) multiplied by the cosine of the angle of the ramp.

I hope this helps! Remember to always draw a diagram and label all the given information to help you visualize and solve the problem. Best of luck!