Question

An inclined plane of angle θ = 20.0° has a spring of force constant k = 495 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 = 2.29 kg is placed on the plane at a distance d = 0.297 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?

Answer 1

  We have no information about friction, so let's assume that it is negligible.
Also assume that the angle of the slope is measured from the horizontal.
Assume ideal spring

The initial kinetic energy of the block plus the change in potential energy will equal the spring potential energy at maximum compression

KE + PE = PS

½mv² + mgh = ½kx²
mv² + 2mgh = kx²

h = (d + x)sinθ

mv² + 2mg(d + x)sinθ = kx²

0 = kx² - 2mgxsinθ - mv² - 2mgdsinθ

which is a quadratic equation in x

0 = 495x² - 2(2.29)(9.81)xsin20 - (2.29)0.750² - 2(2.29)(9.81)0.297sin20
0 = 495x² - 15.366x - 3.275

x = (15.366 ±√(15.366² - 4(495)(-3.275))) / (2(495)

x = 0.0983 m
or
x = -0.0672 m (as a negative value represents a spring stretch instead of compression, we ignore this result)

x = 0.0983 m ANSWER