# Solving Spring Slope Troubles w/ Frictionless Slope

• M1ZeN
In summary, a 0.5 kg book is compressed 0.5m by a spring at the top of a 30 degree slope. The spring constant is 25 kg/s^2. When the book is released, it will reach the edge of the slope at the spring's equilibrium point and then start sliding down the slope. The speed of the book after it has traveled 5 meters in the x direction on a frictionless slope can be determined using energy equations. If the book's mass is doubled, the answer for part (a) will change, but by how much is still unclear.
M1ZeN

## Homework Statement

A spring at top of a 30 degree slope is compressed 0.5m by a 0.5 kg book. If the book is released, it will just reach the edge of the slope at the spring's equilibrium point, and then it will start to slide down the slope. Assume the spring constant is 25 kg/s^2. Assume X-direction and Y-direction mean left-right, up-down as usual.

a) If the slope is frictionless, what will be the speed of the book after it has traveled a distance 5 meters in the x direction?

b) IF the book's mass is doubled, does your answer for (a) change, and if so, by how much (b)?

Fsp = -k(s)

## The Attempt at a Solution

I have made a diagram of what the problem is trying to make me visualize. I understand physically what is going on but however I don't how to start off with a right equation. There is just too many unknowns that is given. Part B is straightforward, it's just Part A I'm running myself into a wall.

Same as in the other spring problem you just posted. Check out the energy equations.

I would approach this problem by first identifying the relevant equations and principles that can be used to solve it.

The first equation that comes to mind is Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This can be expressed as Fsp = -kx, where Fsp is the force exerted by the spring, k is the spring constant, and x is the displacement.

In this problem, the book is released at the equilibrium point of the spring, meaning that the force exerted by the spring is equal to the weight of the book (mg). We can set these two forces equal to each other and solve for x to find the displacement of the spring:

mg = kx

x = mg/k

Since we know the mass of the book (0.5 kg) and the spring constant (25 kg/s^2), we can find the displacement of the spring to be x = 0.02 m.

Next, we can use the equations of motion to solve for the velocity of the book after it has traveled 5 meters in the x-direction. Since the slope is frictionless, we can assume that there are no external forces acting on the book except for gravity. Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration (equal to g = 9.8 m/s^2), and s is the displacement, we can solve for v to find the final velocity of the book after it has traveled 5 meters in the x-direction:

v = √(2as) = √(2*9.8*5) = 9.9 m/s

Therefore, the speed of the book after it has traveled 5 meters in the x-direction is 9.9 m/s.

For part b, if the mass of the book is doubled, the force exerted by the spring (mg) will also double, but the spring constant and displacement remain the same. This means that the final velocity (v) will also double, as all other variables in the equation remain constant. Therefore, the answer for part a would change to v = 19.8 m/s.

In conclusion, by using the relevant equations and principles, we can

## 1. What is a frictionless slope?

A frictionless slope is a theoretical slope that has no resistance or friction, allowing objects to slide down without any external forces acting upon them.

## 2. How does friction affect the motion of an object on a spring slope?

Friction can slow down or stop the motion of an object on a spring slope, as it creates resistance against the movement of the object. This can affect the accuracy of the calculations for solving the spring slope troubles.

## 3. What is the relationship between the angle of the slope and the force exerted by the spring?

The steeper the slope, the greater the force exerted by the spring. This is because a steeper slope means that the object is traveling a longer distance, causing the spring to stretch further and exert a greater force.

## 4. How can one account for friction in solving spring slope troubles?

To account for friction, one can use the coefficient of friction, which is a measure of the amount of friction between two surfaces. This value can be used in calculations to adjust the force exerted by the spring and the acceleration of the object on the slope.

## 5. Can a spring slope be completely frictionless in real life?

No, a completely frictionless slope is not possible in real life as there will always be some amount of friction present between surfaces. However, the amount of friction can be minimized through the use of smooth materials and lubricants, allowing for a more accurate simulation of a frictionless slope.

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