k x = 49 N
1/2 k x^2 = 1/2 m v^2 ... k x^2 = m v^2 = .98 kg * 1.44 m^2/s^2 = 1.41 N⋅m
x = (k x^2) / (k x) = (1.41 N⋅m) / 49 N = .029 m
k = k x / x = 49 n / .029 m
: A 0.98-kg block is held in place against the spring by a 49-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 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 0.32. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops.what is the spring constant of the spring
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