An inclined plane of angle theta = 20. 0 degree has a spring of force constant k

Question

An inclined plane of angle theta = 20. 0 degree has a spring of force constant k = 525 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. 61 kg is placed on the plane at a distance d = 0. 288 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? differs from the correct answer by more than 10%. Double check your calculations. m

Explanation / Answer

using energy conservation,

mu2 /2 + mg(0.288+x)sin20 = kx2 /2

2.61 x 0.752 /2 + 2.61 x 9.81 x (0.288+x)sin20 = 525x2 /2

0.734 + 2.52 + 8.76x = 262.5x2

262.5x2 - 8.76x - 3.254 = 0

x = 0.13

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