Question

Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside the gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance xc while inside of the gun). Use g for the magnitude of acceleration due to gravity. Express the distance L in terms of xc, k, m, g, μ, and θ.

Answer 1

Answer:

L = [0.5*k*(Xc)^2 - m*g*Xc*sin(θ)] / [m*g*μ*cos(θ) + m*g* sin(θ)]

Explanation:

To understand this problem we must analyze well the description of this problem, that is why we went to google, to look for a clearer description of this as well as the question.

In the attached image we find, a graphical description of the problem, very important to be able to perform an analysis.

On the Internet was founded the widest description and the same for the question.

Full description:

A block of mass m is placed in a smooth-bored spring gun at the bottom of the incline so that it compresses the spring by an  amount Xc. The spring has spring constant k. The incline makes an angle θ with the horizontal and the coefficient of kinetic

friction between the block and the incline is μ. The block is released, exits the muzzle of the gun, and slides up an incline a  total distance L.

Question

Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside  the gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance Xc while  inside of the gun). Use g for the magnitude of acceleration due to gravity.

Now we begin to analyze the problem with the data they give us, so we can find the distance L in terms of Xc, k, m, g, μ, and θ.

We will use the theorem of work and energy conservation, which tells us that the initial mechanical energy plus the work done on the body must be equal to the final mechanical energy.

Therefore:

$E_{1}+W_{1-2}=E_{2}$

where

E1 = initial energy [J]

W1-2 = trabajo realizado [J]

E2 = final energy [J]

We will use as a reference point of potential energy, at the moment the mass m is compressed before leaving the gun, ie at this point the potential energy is zero.

$E_{1}=E_{pot}+E_{kin}+E_{elas}\\E_{1}=0+0+E_{elas}$

We notice that the kinetic energy at first is equal to zero because there is no movement.

Therefore:

$E_{1}=E_{elas}\\E_{1}=\frac{1}{2} *k*x_{c}^{2}$

In order to find the work done by friction we must perform a scheme of the block moving on the inclined plane, with the forces acting in the direction of orientation of the plane. In another attached image we can see this diagram.

We know that the friction force is equal to the product of normal force by the coefficient of friction.

f = N * μ

where:

f = friction force [N]

N = normal force [N]

μ = friction coefficient

We then perform a sum of forces equal to zero on the Y axis, since there is no movement on the Y axis.

-m*g*cos(θ) + N = 0

N = m*g*cos(θ)

Therefore:

f = m*g*cos(θ)*μ

Here we must understand that the friction force is doing work in the opposite direction to the movement, therefore the work is negative.

W1-2 = - L*m*g*μ*cos(θ)

Energy at the end, we can easily deduce to interpret the block when it reaches the maximum position, has traveled the distances L & Xc in the inclined plane, at this point, its speed is zero, therefore there is no kinetic energy and no interaction of the spring, therefore its elastic energy is zero. The only energy that exists is the potential, and we must find the height in function of the rectangle triangle that forms.

E2 = m*g* (Xc + L)*sin(θ)

Now we organize the terms:

0.5*k*(Xc)^2 - L*m*g*μ*cos(θ) = m*g* (Xc + L)*sin(θ)

Clearing the variable L, we have:

0.5*k*(Xc)^2 - L*m*g*μ*cos(θ) = m*g*Xc*sin(θ)+ m*g*L*sin(θ)

0.5*k*(Xc)^2 - m*g*Xc*sin(θ) =  L*m*g*μ*cos(θ) + m*g*L*sin(θ)

L = [0.5*k*(Xc)^2 - m*g*Xc*sin(θ)] / [m*g*μ*cos(θ) + m*g* sin(θ)]