# PE, Work, object on inclined plane resting on a spring.

• J-dizzal
In summary: You might need a new one.In summary, a block with a mass of 2.66 kg is placed on a frictionless incline with a spring compressed at 19.1 cm and released. The elastic potential energy of the compressed spring is 25.53 J and the change in gravitational potential energy for the block-Earth system as it moves from the release point to the highest point on the incline is -15.69 N. The highest point is located 0.612 m from the release point.
J-dizzal

## Homework Statement

A block with mass m = 2.66 kg is placed against a spring on a frictionless incline with angle θ = 37.0° (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 14 N/cm, is compressed 19.1 cm and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_43.gif

## Homework Equations

W=Fd, U(x)=1/2 kx2, U(y)=mg(Δy)

## The Attempt at a Solution

Im on part b, I am trying to solve U(y) but i don't have distance traveled in y direction. I think i need to find velocity next.
[/B]

Are you incapable of doing something or do you want to see someone else doing it?

24forChromium said:
Are you incapable of doing something or do you want to see someone else doing it?
Its more of a combination of both of those, but I am looking for help to understand the technique in solving problems.
This being my first semester of physics, I've noticed that solving physics problems is like working on a car, the first time you do something it takes 10 times longer than the second. All the problems I've posted on here are different concepts than what i have posted before.

24forChromium said:
Are you incapable of doing something or do you want to see someone else doing it?
I have a feeling you know of an obvious solution to this problem, do you have any hints?

In part a) it says 19.1cm, but in your working you rounded it to 19cm.
Your notes to the right of that show F=W/d. That only works for a constant force acting over distance d. If you want to know the compression force in the spring (but you don't need it for this question), use F=kx.
For part b), think about energy, and how you can use the result from part a).

haruspex said:
In part a) it says 19.1cm, but in your working you rounded it to 19cm.
Your notes to the right of that show F=W/d. That only works for a constant force acting over distance d. If you want to know the compression force in the spring (but you don't need it for this question), use F=kx.
For part b), think about energy, and how you can use the result from part a).
for part c, I am getting a distance of .612m, which is the distance from the release point along the inclined plane of which the object traveled. Using the formula U(x) =mg(x-x0) and solving for (x-x0). where, U(x)=25.53J, m=2.66kg, g=-15.69N

J-dizzal said:
g=-15.69N
Why that value, and how come units of Newtons?

haruspex said:
Why that value, and how come units of Newtons?
because its the component of weight that alone the axis of the plane.

J-dizzal said:
because its the component of weight that alone the axis of the plane.
Then you mean ##F=mg\sin(\theta)=-15.69N##. But how does that lead to .612m?

haruspex said:
Then you mean ##F=mg\sin(\theta)=-15.69N##. But how does that lead to .612m?
I thought it would make more sense to use it instead of -9.8m/s/s because the object is on an incline. I tried plugging in 9.8 but that still gives the wrong value. I tried using the kinematic displacement formula but time is not known. I have the objects initial velocity at release but I am not sure how to get distance from that.

haruspex said:
Then you mean ##F=mg\sin(\theta)=-15.69N##. But how does that lead to .612m?
i don't see why this wouldn't work;
W=Fd
d=W/F
d=(-25.53)/(-15.69)
d=400.6m​

J-dizzal said:
I thought it would make more sense to use it instead of -9.8m/s/s because the object is on an incline.
By all means use ##g\sin(\theta)## instead of g, but don't write "g=" that value. g doesn't change. You compounded the error by incorporating the mass, turning it from an acceleration into a force. That's why I suggested you meant "F=". All very confusing.
J-dizzal said:
d=(-25.53)/(-15.69)
d=400.6m​
,
Your calculator appears to be broken.

J-dizzal

## What is the relationship between potential energy and work?

Potential energy is the stored energy an object has due to its position or configuration. Work is the transfer of energy from one object to another. The relationship between potential energy and work is that when an object moves from a higher potential energy state to a lower one, work is done and energy is transferred.

## How does an object's position on an inclined plane affect its potential energy?

An object's potential energy is directly related to its height or position above the ground. On an inclined plane, the higher the position of the object, the greater its potential energy. This is because the object has more gravitational potential energy due to its height above the ground.

## How does a spring affect the potential energy of an object on an inclined plane?

A spring is a type of elastic potential energy that can transfer energy to an object when compressed or stretched. When an object on an inclined plane is resting on a spring, the spring's potential energy is transferred to the object and adds to its overall potential energy.

## What factors affect the potential energy of an object on an inclined plane resting on a spring?

The potential energy of an object on an inclined plane resting on a spring is affected by the mass of the object, the height of the inclined plane, and the compression or stretch of the spring. The greater the mass, height, or compression/stretch, the greater the potential energy.

## How can potential energy and work be used to calculate the force required to move an object on an inclined plane resting on a spring?

The force required to move an object on an inclined plane resting on a spring can be calculated using the equations for potential energy and work. First, the potential energy of the object at its initial position must be calculated. Then, the change in potential energy due to the object's movement down the incline can be calculated. This change in potential energy is equal to the work done on the object by the force of gravity. By setting these two equations equal to each other, the required force can be solved for.

Replies
27
Views
6K
Replies
9
Views
2K
Replies
2
Views
2K
Replies
15
Views
3K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
11
Views
3K
Replies
21
Views
4K
Replies
17
Views
4K
Replies
1
Views
1K