An 0.80-kg block is held in place against the spring by a 67-N horizontal extern

Question

An 0.80-kg block is held in place against the spring by a 67-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v1=1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v2=1.9 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.39. The velocity of the block is v3=1.4 m/s at C. The block moves on to D, where it stops. The spring constant of the spring is closest to

Explanation / Answer

by conservation of mechanical energy

(1/2)m(V_1)^2 = F*x

(1/2)*0.8*(1.2)^2 = 67*x

x = 0.00859701 m = 8.597 mm

now we know that

F = kx

so

67 = k * 0.00859701

k = 7793.41 N/m

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