Force of Friction: 25 kg Block Up Ramp 20m Long 3m High

[rjulius]

Homework Statement

A 25 kg block is pulled up a ramp 20 meters long and 3 meters high by a constant force of 120 N. If the box starts from rest and has a speed of 2 m/s at the top, what is the force of friction between the box and the ramp?

Homework Equations

The Attempt at a Solution

 

[SammyS]

Use the work-energy theorem:

The work done on a body by the net force is equal to the change in its kinetic energy.​

 

[Dickfore]

Use the work-energy theorem:

The work done on a body by the net force is equal to the change in its kinetic energy.​

Or, the work done by all the non-conservative forces is (negative) equal to the change in the total mechanical energy (kinetic + potential energy). Then, work done by a constant force is the product of the force and the displacement along the line in which the force is acting. If the angle is obtuse (opposite directions), then this work is negative.

 

[rjulius]

please in formula? I can't understand the question coz i hadn't got any sleep yet sorry

 

[Ashu2912]

Use the work-energy theorem : The work done by all forces on an object = change in it's kinetic energy.
(1) You've got the final and initial velocities, so you can get the change in kinetic energy.
(2) All the forces here are constant, so you can directly use F.s
(3)The forces are the applied force, friction, weight and normal reaction.
(4) Choose the coordinate system, preferably, with the an axis on the slope. You've also got the angle of the slope.
(5) You know the displacement. So, find out the work done by each force by using F.s = F.s.cos(angle between the F and s), and equate it to the change in kinetic energy.
You have one unknown, the force of friction, so one equation will take you through!

 

[rjulius]

use the work-energy theorem : The work done by all forces on an object = change in it's kinetic energy.
(1) you've got the final and initial velocities, so you can get the change in kinetic energy.
(2) all the forces here are constant, so you can directly use f.s
(3)the forces are the applied force, friction, weight and normal reaction.
(4) choose the coordinate system, preferably, with the an axis on the slope. You've also got the angle of the slope.
(5) you know the displacement. So, find out the work done by each force by using f.s = f.s.cos(angle between the f and s), and equate it to the change in kinetic energy.
You have one unknown, the force of friction, so one equation will take you through!

thank you very much for helping me, now i can sleep.

 

[Ashu2912]

Good night. What's the answer by the way?

 

[rjulius]

hehehe I am just answering it, but i think ill be over in no time . ill just post the answer right after.

thanks

 

[rjulius]

Good night. What's the answer by the way?

im skeptical but i found out that its 2343.779092


seems big for a friction force? . . . is it?

 

[Dickfore]

im skeptical but i found out that its 2343.779092


seems big for a friction force? . . . is it?

This is not an acceptable result for a force.

 

[rjulius]

This is not an acceptable result for a force.

yeah you're right urrrghh ill just re-compute it when i wake up this morning, the dawn is getting over my brains.

may be a few hours or sleep would wake my brain up, well thanks for noticing my mistake. good day!

 

[PeterO]

Homework Statement

A 25 kg block is pulled up a ramp 20 meters long and 3 meters high by a constant force of 120 N. If the box starts from rest and has a speed of 2 m/s at the top, what is the force of friction between the box and the ramp?

Homework Equations

The Attempt at a Solution

Alternative 1.
The body will gain a certain amount of Potential energy - it is higher in elevation at the top of the slope than at the bottom.
The body has gained a certain amount of Kinetic energy - it is moving faster when it reaches the top.
In the absence of friction, what force would you need to achieve that? - let's say it was 85 N.
The friction Force must be the difference between the actual applied force and the required force - so if 85 was correct - friction would be 35N.

note: if when you work it out, 85n is the right interim answer, you may toast my guess - I just made up a number that was less than 120!

Alternative two:

Energy is added to the block when you apply the 120N Force along the full length of the slope.

Energy is taken from the block by the friction force, which also acts along the full length of the slope.

The gain in energy the block has is partly because it is higher up [mgh] and partly because it is now moving at a speed [1/2mv^2]

you can easily calcuate the Work done by the applied force, the Kinetic Enegy gained and the potential energy gained. From that you can calculate how much work is done by friction - and therefore the size of the Friction Force.

Peter

 

[Redbelly98]

Moderator's note:

im skeptical but i found out that its 2343.779092


seems big for a friction force? . . . is it?

rjulius: please show your work.

Others: please wait for rjulius to show his work before offering further help.

 

[Dickfore]

yeah you're right urrrghh ill just re-compute it when i wake up this morning, the dawn is getting over my brains.

may be a few hours or sleep would wake my brain up, well thanks for noticing my mistake. good day!

Did you get enough sleep already? You never posted the correct solution.