Inclined Plane with spring. Show your work and explain your reasoning. An inclin

Question

Inclined Plane with spring. Show your work and explain your reasoning. An inclined plane of angle theta=25.0 degrees has a spring of force constant k=515 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure. A block of mass m=2.45kg is placed on the plane at a distance d=0.319 m from the spring. From this position the block is projected downward toward the spring with the speed v=0.730 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Explanation / Answer

the statement of enegy conservation:

initially the energy of the block is its kinetic energy and the gravitaional potential enegy and finally it has spring potential energy

since the block is compresses by x distance in final position so it is at a distance of (d+x) from original position...thus at a hight of (d+x)*sin(theta) from initial position

so gravitaional potential enegy=m*g*(d+x)*sin(theta)

kinetic energy=0.5*m*v^2

spring potential energy=0.5*k*x^2

thus by energy conservation

mg*(d+x)*sin(theta)+0.5*m*v^2=0.5*K*x^2

=> 257.5x^2-10.147x-3.8897=0

solving the quadratic we get

x=0.144 m=14.4 cm=====> compression in the spring

the other root of the quadratic is neglected because it comes out to be negetive and by defenition compression in the spring can not be negetive

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