Question

An inclined plane of angle 0 = 20.0° has a spring of force constant k = 500 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in Figure P7.63. A block of mass m = 2.50 kg is placed on the plane at a distance d = 0.300 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? Figure P7.63

Answer 1

Using law of conservation of energy,

0.5 mv^2 + mg d sin(theta) = 0.5 kx^2

(0.5 x 2.5 x 0.75^2) + (2.5 x 9.8 x (0.3 + x) sin(20)) = 0.5 x 500 x^2

250 x^2 - 8.38 x - 3.22 = 0

X= 0.131 m

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