A rock (0.4 kg) sits on a spring (k = 1000 N/m) and expression is 0.3 m (a) What

Question

A rock (0.4 kg) sits on a spring (k = 1000 N/m) and expression is 0.3 m (a) What is the maximum height the rock will reach it is launched from the spring? How fast will the rock be traveling just before it hits the ground? (Assume the spring has been removed) (c) If the rock hits a stick and drives the stick 0.1 m into the ground, with what force did the rock hit? (d) Is the work done on the stick greater than, less than, or equal to the potential energy initially stored in the spring? Explain your answer.

Explanation / Answer

a)

x = compression of spring = 0.3 m

m = mass = 0.4 kg

k = spring constant = 1000 N/m

using conservation of energy

potential energy gained by rock at height "h" = elastic potential energy lost

mgh = (0.5) k x2

inserting the values

(0.4) (9.8) h = (0.5) (1000) (0.3)2

h = 11.5 m

b)

using conservation of energy between Top and Bottom again

Kinetic energy just before hitting the ground at bottom = Potential energy at the top

(0.5) m v2 = mgh

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 11.5)

v = 15.01 m/s

c)

work done on the stick = kinetic energy of the rock

F d = (0.5) m v2

F (0.1) = (0.5) (0.4) (15.01)2

F = 450.6 N

d)

the work done on the stick is same as the energy stored by the spring initially as per conservation of energy which says that energy only converts from one type to another type.

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