A block of mass m is placed in?

Question

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 x_c. The spring has spring constant k. The incline makes an angle (theta) with the horizontal and the coefficient of kinetic friction between the block and the incline is mu. The block is released, exits the muzzle of the gun, and slides up an incline a total distance L.

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 x_c while inside of the gun). Use for the magnitude of acceleration due to gravity.
Express the distance L in terms of x_c, k, m, g, mu, and (theta).

Answer 1

We can do this problem using forces and energy..

Let's take a moment to look at all of the pieces to our puzzle.

The spring has potential energy, given by
Ps = ½ k (x_c)²

Friction does work against the motion of the block, at a rate of
Wf = -µ N L

Wf = Work done by friction
µ = coefficient of friction
N = normal force (which we will need to get from Newton's laws)
L = total distance traveled.



The block has an initial speed of zero, so KEi = 0 and a final speed of zero, so KEf = 0.



The first thing we need is the value for N. If you draw a free-body diagram, you will see that
N = mg cos θ


Draw a right triangle. L is the hypotenuse, h is the height above the floor the block travels, θ is the base angle. It's easy to see that h = L sin θ.


We are dealing with friction, which is a non-conservative force. Therefore the work done by friction is just the change in energy. As stated earlier, the only energy involved is the potential energy due to the spring (initially) and the height the block rises (finally)

Wf = Ef - Ei

-µ N L = mgh - ½k (x_c)²

-µ mg L cos θ = mg L sin θ - ½k (x_c)²

mg L sin θ + µ mg L cos θ = ½k (x_c)²

L mg (sin θ + µ cos θ) = ½k (x_c)²

L = k (x_c)² / (2mg [sin θ + µ cos θ])



There you go, I hope that helps.

Answer 2

To find L, you have to find the distance traveled along the incline by the block after it exits the gun. You have to ignore friction when the block is inside the gun. You also have to assume that the uncompressed spring is just at the top of the gun. Using g as the magnitude of acceleration due to gravity, it equals 9.8 meters/sec.
Please let me know if this helped you. I am working on getting my bachelors in art history from MIT to help educate people from the dominican republic like Karl how to best take care of their pets like cats and dogs.
Therefore, the answer is 7.5m*0.2g and theta equals 100 degrees.

Answer 3

if you enter the answer above it'll tell you that Your answer would be correct if L were measured from the initial position of the block, but you need to take into account that the block starts a certain distance inside the gun.
the correct answer is
(.5k(x_c^2)-mgsin(theta))/(mg(sin(theta)+mu*cos(theta)

Answer 4

this is pretty sucky problem it takes a relatively easy concept and makes it hard with extra busy work
I wont give you the answer but you can im me at guitaristicny on yahoo if you are really stuck.

Basically this is a problem about transformation of engery. the spring is compressed which gives it a potential energy. When it is released the block is given a potential energy and slides up the incline. the energy of the block is partially transformed into gravitational potetnial energy and partially turned into heat from friction.


this is to get you started. the initial energy of the block is gonna be (1/2)k*x^2which is going to equal the energy lost to friction plus the gravitational potential energy at the top.

one more hint, dont let the angle part confuse you it doesnt play that big a role

Answer 5

L = 0.5*(xc)^2*(k)−(m*g*sin(θ)*xc/mg(sin(θ)+cos(θ)μ)