An inclined plane of angle = 20.0° has a spring of force constant k = 450 N/m fa

Question

An inclined plane of angle = 20.0° has a spring of force constant k = 450 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.41 kg is placed on the plane at a distance d = 0.291 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

An inclined plane of angle = 20.0° has a spring of force constant k 450 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m 2.41 kg is placed on the plane at a distance d 0.291 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Explanation / Answer

Using Work-energy theorem

Wnet = dKE

Wg + Ws = KEf - KEi

Fg*(d + x) + (0.5*k*xf^2 - 0.5*k*xi^2) = 0.5*m*vf^2 - 0.5*m*vi^2

xf = 0, and vf = 0

and xi = x, and vi = v

m*g*sin A*(d+x) - 0.5*k*x^2 = 0 - 0.5*m*v^2

using given values

2.41*9.81*sin 20 deg*(0.291 + x) - 0.5*450*x^2 + 0.5*2.41*0.75^2 = 0

8.086*0.291 + 8.086*x - 225*x^2 + 0.6778 = 0

225x^2 - 8.086*x - 1.6752 = 0

Now solving above quadratic equation:

x = [8.086 +/- sqrt (8.086^2 + 4*1.6752*225)]/(2*225)

taking positive sign

x = [8.086 + sqrt (8.086^2 + 4*1.6752*225)]/(2*225)

x = 0.106 m

Let me know if you have any doubt.

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