# Distance of box pushed up incline by spring

• anothertangent
In summary, the problem involves a block of mass 1.5kg attached to a spring with constant 590N/m on an inclined plane with an initial angle of 21 degrees and coefficient of kinetic friction of 0.19. The block is initially at rest with the spring compressed by 0.18m and the initial gravitational energy is 0. The task is to find the distance L that the block will travel as the spring pushes it up the incline. Using the equations for work, kinetic energy, potential energy, and kinetic friction, the solution involves finding the balance between the initial mechanical energy of 9.558J and the energy spent on potential energy and overcoming friction until the block stops. This can be achieved by
anothertangent

## Homework Statement

A block of mass m=1.5kg is attached to a spring of constant k=590N/m. It is initially at rest on inclined plane of 21 degrees, and coefficient of kinetic friction between block and plane is MUk=.19. In initial position, where spring is compressed by d=.18m, mass is at lowest position and spring is compressed maximum amount. Take initial gravitational energy of block as 0. If spring pushes the block up the incline, what distance L will the block travel, in m?

## Homework Equations

W=Fd
KE=.5mv^2
PE=mgh
PE of spring=.5kx^2
Fkinetic friction= MUk(mgcos(theta))
KE0+PE0+Wfriction=KE+PE

## The Attempt at a Solution

I assume the basic formula to be KE0+PE0+Wfriction=KE+PE. I have solved for the initial mechanical energy in the first part of the problem, and it equals 9.558J. The equation then becomes 9.558+Wf=KE+PE. Because it's a spring, the PE=.5kx^2, so the equation becomes 9.558+Wf=.5mV^2+.5kx^2, and because the block's final speed is 0, the equation is 9.558+Wf=.5kx^2. Substituting for the spring constant becomes 9.558J+Wf=.5(590N/m)x^2. I assume that if I could solve for x, I could calculate the total distance traveled, as the spring started at being compressed 18m. Wf should equal Ffriction times L, and I calculated 2.607N for Ffriction, from Fk=MUk(mgcos(theta)). I'm a bit unsure of where to go next, as I seem to have hit a wall; I don't know Wf, x (which equals L-.18), or L (which I'm trying to solve for). Any help is greatly appreciated.

You have correctly found that the block starts with 9.558 J worth of energy. It spends these Joules (a) to buy potential energy and (b) to overcome friction until it stops. Both quantities involve the distance L up the incline. Can you write (a) expressions for each and (b) the energy balance equation?

## 1. How does the distance of a box pushed up an incline by a spring relate to the spring constant?

The distance of a box pushed up an incline by a spring is directly proportional to the spring constant. This means that as the spring constant increases, the distance the box is pushed up the incline also increases.

## 2. What is the formula for calculating the distance of a box pushed up an incline by a spring?

The formula for calculating the distance of a box pushed up an incline by a spring is d = (1/2) kx^2, where d is the distance, k is the spring constant, and x is the displacement of the spring.

## 3. Can the angle of the incline affect the distance the box is pushed up by the spring?

Yes, the angle of the incline can affect the distance the box is pushed up by the spring. The steeper the incline, the shorter the distance the box will travel as it is pushed up by the spring.

## 4. How does the mass of the box affect the distance it is pushed up by the spring?

The mass of the box does not directly affect the distance it is pushed up by the spring. However, a heavier box may require a stronger spring with a higher spring constant to achieve the same distance.

## 5. Is the distance of the box pushed up by the spring affected by external forces?

Yes, external forces such as friction and air resistance can affect the distance the box is pushed up by the spring. These forces can cause the box to lose energy and decrease the distance it travels up the incline.

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