4) (13 pts.) A block of mass m is placed in a smooth-bored spring gun at the bot

Question

4) (13 pts.) 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 arc. The spring has spring constant k. The incline makes an angle with the horizontal and the coefficient of kinetic friction between the block and the incline is A. 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, A, and 6. Hint: Pay attention to your sign! (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 ae inside of the gun

Explanation / Answer

(A)

initial energy of the block=potential energy of the spring

=0.5*spring constant*compression^2

=0.5*k*xc^2

(B):

In this the formula usese.....

friction force=friction coefficient*normal force

=mu*m*g*cos(theta)

as friction force opposes the motion, angle between force and displacement is 180 degrees

hence work done=force*distance*cos(180)

=-mu*m*g*cos(theta)*L

(C)

In this the ...

final energy of the block=potential energy of the block

=mass*g*height from zero potential energy position

=m*g*(L+xc)*sin(theta)

(D)

using energy conservation principle:

initial energy+work done=final energy

=0.5*k*xc^2-mu*m*g*cos(theta)*L=m*g*(L+xc)*sin(theta)

==>0.5*k*xc^2-m*g*xc*sin(theta)=m*g*L*sin(theta)+mu*m*g*cos(theta)*L

==>L=(0.5*k*xc^2-m*g*xc*sin(theta))/(m*g*sin(theta)+mu*m*g*cos(theta))

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