Work done by a friction force, block moving up a ramp-question

[AndreaTechGirl]

Work done by a friction force, block moving up a ramp--question

Homework Statement

100kg block moves up on a rough surface, at an incline of 30 degrees, a constant force P=800 N is applied horizontally moving the block a distance of 3 M up the ramp in a time interval of 2 seconds, v1=0.8 m/s, v2=2.2 m/s, what is the work done by the friction force

Homework Equations

d-displacement, W-work, F-force, m-mass, g-gravity, v1-initial velocity, v2-terminal velocity
eq1) W=F*cos(30)*d
eq2) F(friction)=(coefficient of kinetic friction)*mg
eq3) W(friction)=F(friction)*d*cos(theta)
eq4) 1/2m(v2)^2-1/2m(v1)^2=F(friction)*d*cos(theta)

The Attempt at a Solution

Change in KE of the 8block is equal to the work done on block by friction force so, plugging in for eq4 i get 210, it should be negative because its in the opposite direction-right? Is my set up correct? Thanks for your time!

 

[Astronuc]

There is a force accelerating the block, which means changing the velocity, and friction is not part of that. The force doing the accelerating is the applied force minus the friction force.

The total force includes the force accelerating the block AND the friction force which opposes the movement of the block.

The work done by friction is a function of the friction force applied over the distance.

 

[denverdoc]

(astronuc, good to see you here, simultaneity at work here)

It is all good, though eqn 4 may not be the best choice for solving and I didn't check the results, but I suspect may be in error cause it neglects the slowing that would be present due to work done by/ or against gravity. Either way 4 is best when there is no change in Potential Energy. I like 3 here.

PS: this may be handy and signs can be awfully confusing in work.
Work sign rule is : if component of force is in the direction of displacement, then work is positive, otherwise negative.
Positive work means transfer of energy "to" the particle.
Negative work means transfer of energy "from" the particle.

 

[AndreaTechGirl]

There is a force accelerating the block, which means changing the velocity, and friction is not part of that. The force doing the accelerating is the applied force minus the friction force.

The total force includes the force accelerating the block AND the friction force which opposes the movement of the block.

The work done by friction is a function of the friction force applied over the distance.

Wow! that was fast! Thank you!

Is the applied force P=800 N? For the accelerating force do I use the F=ma? Do I use F=m ((v2)^2-(v1)^2)/(2d)? so I get 730 for my friction force?? So confused!

 

[AndreaTechGirl]

Thank you for that tip, denverdoc! I am trying to figure out a way to use eq3, but I don't have the friction force, that's what I am solving for, but I have velocities and a mass, which is why I chose to use eq4

 

[Feldoh]

You might want to look at equation 2 again -- I think it should be $$F_f = \mu_kF_N$$

 

[denverdoc]

You might want to look at equation 2 again -- I think it should be $$F_f = \mu_kF_N$$

Indeed, it is. And in this case critical to soln as the normal force on a level surface= mg and is less on an an incline.

 

[denverdoc]

Thank you for that tip, denverdoc! I am trying to figure out a way to use eq3, but I don't have the friction force, that's what I am solving for, but I have velocities and a mass, which is why I chose to use eq4

Thats good reasoning, but only if you account for the difference in potential energies, after all even in a vacuum, a ball will still slow when thrown upwards,

 

[AndreaTechGirl]

So, taking a different approach--I need to find all forces acting on the block, so if force P=800 in a horizontal direction (incline is 30), then its x and y components are 800N*cos(30), and 800Nsin(30). ANd if its weight force is mg=980N, then its x and y components are 980N*cos(30) and 980*sin(30), and those would both be negative because they are in the direction opposite of the movement, and my friction force is going to be -980sin(30)=-490 N, does that seem right so far?

 

[AndreaTechGirl]

Thats good reasoning, but only if you account for the difference in potential energies, after all even in a vacuum, a ball will still slow when thrown upwards,

Sooo True! THanks for helping me out! I really appreciate it!

 

[denverdoc]

Hey I'm a rocket guy and know all too well what goes up usually comes down, and occasionally in many more pieces, esp without the help of a parachute. I put a lot of work into it, and nowhere in the work-energy theorum does that get accounted for. Universe you owe me!

 

[AndreaTechGirl]

I found the friction force to be -133 N, to find work, I multiply the force by the distance, do I need to account for the angle of the incline?

 

[denverdoc]

Not unless there are other energetic terms to account for.

Pe1+KE1=PE2+KE2+frictonal work

edtorial: and VIPoint, some forces like friction burn energy, while others do not, big difference as those that are conservative do not(electric fields, gravity,etc) store or release energy, The conservative forces are like a freeway, while the frictional forces more like a tollway.