A block of mass m is placed in a smooth-bored spring gun at the bottom of the 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_e. The spring has sprung 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. (a) Taking the gravitational potential energy to be zero before the spring is released, what is the initial energy of the block? (b) What is the work done by friction on the block? Ignore friction when the block is inside the gun. Express your answer in terms of L, g, m, mu, and theta. (c) Find an expression for the final energy of the block (the energy when it has traveled a distance L up the incline). Assume that the gravitational potential energy of the block is zero before the spring is released and that the block moves a distance x_0 inside of the gun. (d) Use energy conservation to 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_e while inside of the gun). Express your answer in terms of X_e, k, m, g, mu, and theta.

Explanation / Answer

(a)
Initial Energy = Spring Potential Energy = 1/2*k*xe^2

(b)
Work done by friction, = F*d
W = - u*m*g*cos() * L

(c)
Final Energy of the block = Gravitational Potential Energy
Total distance travelled along the incline = xe + L
Height gained = ( xe + L ) * sin()

Final Energy of the block = m*g* ( xe + L ) * sin()

(d)
Using Energy conservation,
Final Energy = Initial Energy + Work done against friction
m*g* ( xe + L ) * sin() =  1/2*k*xe^2 - u*m*g*cos() * L
m*g* xe * sin() + m*g*L* sin() + u*m*g*cos() * L = 1/2*k*xe^2
L = ( 1/2*k*xe^2  - m*g* xe * sin() ) / (m*g* sin() + u*m*g*cos() )

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