Work Required to Move 890 N Crate Along Floor Against 180 N Force

[tigerwoods99]

Homework Statement

A 890 N crate rests on the floor.
How much work is required to move it at constant speed 5.6 m along the floor against a friction force of 180 N?
How much work is required to move it at constant speed 5.6 m vertically?

Homework Equations

KE = (1/2)mv^2

The Attempt at a Solution

(1/2)(890/9.8) = x?

 

[Phyisab****]

You don't know the velocity so kinetic energy won't help you. Any other equations you can think of?

 

[tigerwoods99]

Work force done = FaDcos(theta)
Where:
Fa = applied force
D = displacement
Cos(theta= I am not sure would it be pie/2

 

[Phyisab****]

cos(pi/2) = 0, here your work is being done in the same direction as the motion so.

That equation gets you half way there, need one more equation.

 

[tigerwoods99]

I think cos(pi/2) = 1.?
Mhm is the other equation:
Work done by friction = FfD
Where Ff= friction force
D = displacement?

 

[Phyisab****]

Actually I wasn't quite right all you need is the equation you just wrote sorry bout that. So just use the equation you just wrote, (W done by friction) = (F of friction)*d
Then just plug and chug.

 

[tigerwoods99]

Thanks I was able to answer part 1.
How about:
How much work is required to move it at constant speed 5.6 m vertically?

So here there is a gravitational component?

 

[Phyisab****]

W=f*d, you know the distance. Can you calculate the force due to gravity?

 

[tigerwoods99]

The force due to gravity is just mg, (860)
W = 890*5.6?

 

[Phyisab****]

there you go easy right? In general if you have an equation with n variables you need n-1 equations to solve for the one variable you're looking for.

 

[tigerwoods99]

Thanks, and thanks for the tip!