An inclined plane of angle = 20.0° has a spring of force constant k = 520 N/m fa

Question

An inclined plane of angle = 20.0° has a spring of force constant k = 520 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 = 2.59 kg is placed on the plane at a distance d = 0.297m 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?

solve and explain please

An inclined plane of angle theta = 20.0 degree has a spring of force constant k = 520 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 = 2.59 kg is placed on the plane at a distance d = 0.297m 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

we can solve it by conservation of energy

0.5mv^2 + mg(h+x) = 0.5kx^2

0.72 +2.59*9.81(0.297cos20 + xcos20) = 0.5*520x^2

7.09 + 0.72+ 23.87 x = 260 x^2

x = 0.225 m

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