Inclined Plane with spring. Show your work and explain your reasoning. An inclin
ID: 2090722 • Letter: I
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