# Calculating work on an inclined plane

• Diana Rico
In summary, you have a problem involving work done on a suitcase being pulled up a ramp at an angle of 25 degrees by a force of 140 N. The coefficient of kinetic friction between the ramp and the incline is given. The solution involves using the equations for work, Newton's Second Law, and finding the components of forces in the x and y directions. Once you have found the work done by each individual force, you can add them together to find the total work done on the suitcase. The final part of the problem involves using the work-energy theorem to find the final velocity of the suitcase after traveling 3.80 m along the ramp.
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.

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 Ffrictionkinetic*FNormal. n is the FNormal, 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

Can you explain Newton's second law?

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

sonnyfab said:
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 Ffrictionkinetic*FNormal. n is the FNormal, 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?

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?

Ced said:
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?

## 1. What is the formula for calculating work on an inclined plane?

The formula for calculating work on an inclined plane is W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical distance the object is moved.

## 2. How does the angle of the inclined plane affect the work done?

The angle of the inclined plane affects the work done by changing the vertical distance the object is moved. The greater the angle, the longer the vertical distance, and therefore the greater the work done.

## 3. What is the role of friction in calculating work on an inclined plane?

Friction plays a role in calculating work on an inclined plane by acting against the motion of the object. It decreases the efficiency of the inclined plane and therefore reduces the amount of work done.

## 4. Can the work done on an inclined plane ever be negative?

Yes, the work done on an inclined plane can be negative if the force of gravity is acting in the opposite direction of the motion of the object. This can occur if the object is moving downhill on the inclined plane.

## 5. What is the relationship between work and energy when calculating work on an inclined plane?

The relationship between work and energy on an inclined plane is that the work done is equal to the change in potential energy of the object. This is because the object is being lifted against the force of gravity, and therefore gaining potential energy as work is done on it.

Replies
7
Views
4K
Replies
56
Views
2K
Replies
24
Views
4K
Replies
2
Views
2K
Replies
41
Views
790
Replies
2
Views
911
Replies
33
Views
2K
Replies
4
Views
1K
Replies
13
Views
504
Replies
4
Views
1K