Calculating work on an inclined plane

[Diana Rico]

A luggage handler pulls a 20.0- kg suitcase up a ramp inclined at 25°above the
horizontal by a force of magnitude 140 N that acts parallel to the ramp. The
coefficient of kinetic friction between the ramp and the incline is If the suit-case
travels 3.80 m along the ramp, calculate
(a) the work done on the suitcase by the force
(b) the work done on the suitcase by the gravitational force;
(c) the work done on the suitcase by the normal force;
(d) the work done on the suitcase by the friction force;
(e) the total work done on the suitcase.
(f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed
after it has traveled 3.80 m along the ramp?

I feel like this problem is so simple yet I cannot get it!

a.) I used W=Frcos(theta) or W=140(3.8)cos(25) but the answer is not correct. The solution says to make theta=90, does that mean that I should use sin in order for it to equal 1 and not 0? The answer is just W=Fr or 532J but why?

b.) I tried to use the same formula because it is the same type of problem but with the angle as 115 degrees and Force as 196 N. W=-315J, but what I do not understand is the proof of why it works which states "Alternatively, the component of w parallel to the incline is wsin25. This component is down the incline so its angle with s is ∅=180°. Wwsin(25)=(196 Nsin25 )(cos180 )(3.80 m)=-315 J. The other component of w, wcos25, is perpendicular to s and hence does no work. Thus Ww=Wwsin(25)=-315J which agrees with the above." Can someone please explain?

c.) I understood W=0 because the normal force and r are perpendicular.

d.) I do not know how the formulas 'n=wcos25' or 'fk=μkn' were obtained to find the answer. Does the the n= have do with the italicized part in part b?

e.) To find the total work done Wt= Wa+Wb+Wc+Wd or Wt= 532J+-315J+0J+(W found in part d) but I do not have part d in order to finish the problem.

f.) For this problem Wt=Kf-Ki or Wt=.5mvf^2-.5mvi^2 and since vi=0 W=.5mvf^2 and v=sqrt(2Wt/20) so all I need is Wt

So really I need help with understanding a, b, and how to get the formula for d please.

 

[sonnyfab]

You're so close

When you do problems with work, you've always got to keep in mind the other physics you know, especially the stuff about forces. You're just missing out on the Newton's Second Law part of this problem and you'll have it solved.

a.) I used W=Frcos(theta) or W=140(3.8)cos(25) but the answer is not correct. The solution says to make theta=90, does that mean that I should use sin in order for it to equal 1 and not 0? The answer is just W=Fr or 532J but why?

The equation for work is W=Frcos(θ) where θ is the angle between the force and the direction of motion. Here, the force is up the ramp, and the motion is up the ramp, so θ is 0.

b.) I tried to use the same formula because it is the same type of problem but with the angle as 115 degrees and Force as 196 N. W=-315J, but what I do not understand is the proof of why it works which states "Alternatively, the component of w parallel to the incline is wsin25. This component is down the incline so its angle with s is ∅=180°. Wwsin(25)=(196 Nsin25 )(cos180 )(3.80 m)=-315 J. The other component of w, wcos25, is perpendicular to s and hence does no work. Thus Ww=Wwsin(25)=-315J which agrees with the above." Can someone please explain?

Awesome. You used the equation and found the angle between the force and the direction of motion. Their solution first found the component of the weight that was parallel to the ramp and then multiplied it by the displacement. The cos (115) and the sin(25) are the same, so your solution using the formula and their solution found by finding the force components are the same.

c.) I understood W=0 because the normal force and r are perpendicular.

Great!

d.) I do not know how the formulas 'n=wcos25' or 'fk=μkn' were obtained to find the answer. Does the the n= have do with the italicized part in part b?

Any force that causes (or prevents) the motion of an object does work on that object (unless is is perpendicular to the direction of travel). You can find the force of friction using the equation F_friction=μ_kinetic*F_Normal. n is the F_Normal, and is found using a free body diagram and Newton's Second Law. If the "x" direction is down the ramp and the "y" direction is perpendicular to the ramp, then gravity has one component that is in the "x" direction and one component is the "y" direction. Since there is no acceleration in the y direction, that part of gravity is equal to the normal force. Then you just multiply that force by the coefficient of friction, , and you know the magnitude of the friction force. Since the friction force points down the ramp and the suitcase moves up the ramp, the angle between them when finding work is 180°.

e.) To find the total work done Wt= Wa+Wb+Wc+Wd or Wt= 532J+-315J+0J+(W found in part d) but I do not have part d in order to finish the problem.

You got it.

f.) For this problem Wt=Kf-Ki or Wt=.5mvf^2-.5mvi^2 and since vi=0 W=.5mvf^2 and v=sqrt(2Wt/20) so all I need is Wt

So really I need help with understanding a, b, and how to get the formula for d please.

I hope that helps. Let me know if I can clarify anything further.

Dr Peter Vaughan
BASIS Peoria Physics

 

[Diana Rico]

Can you explain Newton's second law?

 

[siddharth23]

Newton's second law: F = ma or a = F/m

The acceleration of an object is the ratio of the resultant force acting on it and its mass

 

[Ced]

You're so close

When you do problems with work, you've always got to keep in mind the other physics you know, especially the stuff about forces. You're just missing out on the Newton's Second Law part of this problem and you'll have it solved.

a.) I used W=Frcos(theta) or W=140(3.8)cos(25) but the answer is not correct. The solution says to make theta=90, does that mean that I should use sin in order for it to equal 1 and not 0? The answer is just W=Fr or 532J but why?

The equation for work is W=Frcos(θ) where θ is the angle between the force and the direction of motion. Here, the force is up the ramp, and the motion is up the ramp, so θ is 0.

b.) I tried to use the same formula because it is the same type of problem but with the angle as 115 degrees and Force as 196 N. W=-315J, but what I do not understand is the proof of why it works which states "Alternatively, the component of w parallel to the incline is wsin25. This component is down the incline so its angle with s is ∅=180°. Wwsin(25)=(196 Nsin25 )(cos180 )(3.80 m)=-315 J. The other component of w, wcos25, is perpendicular to s and hence does no work. Thus Ww=Wwsin(25)=-315J which agrees with the above." Can someone please explain?

Awesome. You used the equation and found the angle between the force and the direction of motion. Their solution first found the component of the weight that was parallel to the ramp and then multiplied it by the displacement. The cos (115) and the sin(25) are the same, so your solution using the formula and their solution found by finding the force components are the same.

c.) I understood W=0 because the normal force and r are perpendicular.

Great!

d.) I do not know how the formulas 'n=wcos25' or 'fk=μkn' were obtained to find the answer. Does the the n= have do with the italicized part in part b?

Any force that causes (or prevents) the motion of an object does work on that object (unless is is perpendicular to the direction of travel). You can find the force of friction using the equation F_friction=μ_kinetic*F_Normal. n is the F_Normal, and is found using a free body diagram and Newton's Second Law. If the "x" direction is down the ramp and the "y" direction is perpendicular to the ramp, then gravity has one component that is in the "x" direction and one component is the "y" direction. Since there is no acceleration in the y direction, that part of gravity is equal to the normal force. Then you just multiply that force by the coefficient of friction, , and you know the magnitude of the friction force. Since the friction force points down the ramp and the suitcase moves up the ramp, the angle between them when finding work is 180°.

e.) To find the total work done Wt= Wa+Wb+Wc+Wd or Wt= 532J+-315J+0J+(W found in part d) but I do not have part d in order to finish the problem.

You got it.

f.) For this problem Wt=Kf-Ki or Wt=.5mvf^2-.5mvi^2 and since vi=0 W=.5mvf^2 and v=sqrt(2Wt/20) so all I need is Wt

So really I need help with understanding a, b, and how to get the formula for d please.

I hope that helps. Let me know if I can clarify anything further.

Dr Peter Vaughan
BASIS Peoria Physics

Hi if I may ask, I quite don't understand letter (f). Where did the Wt = Kf - ki come from? Is it from the conservation of energy formula K1 + U1 + Wext = K2 + U2. if yes then how did U1 and U2 become 0? I assume U1 is zero because the suitcase is on the ground but for U2 isn't the suitcase on the plane? so their should ba value for U2 right?

 

[Ced]

Can someone help me with letter (f). Where did the Wt = Kf - ki come from? Is it from the conservation of energy formula K1 + U1 + Wext = K2 + U2. if yes then how did U1 and U2 become 0? I assume U1 is zero because the suitcase is on the ground but for U2 isn't the suitcase on the plane? so their should ba value for U2 right?

 

[PeroK]

Can someone help me with letter (f). Where did the Wt = Kf - ki come from? Is it from the conservation of energy formula K1 + U1 + Wext = K2 + U2. if yes then how did U1 and U2 become 0? I assume U1 is zero because the suitcase is on the ground but for U2 isn't the suitcase on the plane? so their should ba value for U2 right?

This is a very old thread. It's better to post a new thread of your own.