A 968 g block is released from rest at height h0 above a vertical spring with spring constant k = 440 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 22.0 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 6h0 above the spring, what would be the maximum compression of the spring?

a) 10.648 J

b) -10.648 J

c) h₀ = 0.9024 m

d) 0.500 m

Explanation:

The work done by the block on the spring is equal to the work done by the spring on the block

W = (1/2)(k)(x²) = 0.5 (440)(0.22²) = 10.648 J

c) The work done by the block on the spring is the change in potential energy of the block during this fall.

The change in potential energy is

Weight × Total distance fallen through by the block = mg(h₀ + 0.22)

This change in potential energy is now equal to the workdone by the spring on the block

W = mg(h₀ + 0.22)

9.4864 h₀ + 2.087008 = 10.648

h₀ = (8.560992/9.4864)

h₀ = 0.9024 m

d) If the block were released from a height of 6h₀, that is,

H = 6h₀ = (6×0.9024) = 5.415 m above the spring

Work done by the block in falling that height + compression = mg(H + x) = (0.968)(9.8)[5.415 + x)

This work is done on the spring

W = (1/2)(k)(x²)

9.4864[5.415 + x] = 0.5(440)(x²)

51.37 + 9.4864x = 220x²

220x² - 9.4864x - 51.37 = 0

Solving the quadratic equation

x = 0.50 m or - 0.47

The negative answer defies the laws of physics, hence, the decompression of the spring is 0.50 m.

Hope this Helps!!!

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.