with a stiffness of 5,70o N/m and a natural length of 6s A spring re a 40.o kg b

Question

with a stiffness of 5,70o N/m and a natural length of 6s A spring re a 40.o kg block into the air. The springimprestential energy block sitting on top) and then released. (a) How, much potential energy s the stored in the compressed spring? (b) What is the maximum height a hek floor reached by the block? (What is the maximum accelerati atis the (d) At what location does the maximum acceleration occur? ce maximum velocity reached by the block? cm is used to is compressed by 35 cm (with the of the

Explanation / Answer

a) Potential energy stored = 0.5kx^2

= 0.5*5700*0.35^2

= 349.125 J

b) Maximum height reached by the block

0.5kx^2 = mg(h+x)

349.125 = 40*9.8*(h + 0.35)

h = 0.54 m

so the heigth reached above the ground = 0.54 + 0.65 = 1.19 m

c) Net force acting = kx - mg

acceleration = (Fnet) / m

= (5700*0.35 - 40*9.8) / 40

= 40.075 m/s2

d) maximum acceleration occurs at the point of maximum compression when the spring is compresses 35 cm

e) max velocity reached

0.5mv^2 = 0.5kx^2

40*v^2 = 5700*0.35^2

v = 4.18 m/s

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