# Physics question conservation of energy?

Suppose a block is released from rest with the spring compressed 4.70 cm . The mass of the block is 1.60 kg and the force constant of the spring is 990 N/m . Assume the surface is frictionless.

a) What is the speed of the block when the spring expands to a compression of only 2.20 cm?

b)What is the speed of the block after it leaves the spring?

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### 2 Answers

- FiremanLv 72 months agoFavorite Answer
a) The initial energy of the spring i.e. PE(i) = 1/2kx^2

=>PE(i) = 1/2 x 990 x (4.70 x 10^-2)^2 = 1.09 J

b) Let the velocity of the block is v m/s when the spring expands to a compression of only 2.20 cm, by the law of energy conservation:

=>PE(i) of the spring = PE(of spring at this position) + KE(of the block)

=>1.09 = 1/2 x 990 x (2.20 x 10^-2)^2 + 1/2 x 1.60 x v^2

=>v = √1.06

=>v = 1.03 m/s

c) Let the velocity of the block is v m/s when it leaves the spring, by the law of energy conservation:

=>PE(i) of the spring = KE(of the block)

=>1.09 = 1/2 x 1.60 x v^2

=>v = √1.36

=>v = 1.17 m/s

- NCSLv 72 months ago
a) decrease in PE of spring becomes KE:

½ * 990N/m * (0.0470² - 0.0220²)m² = ½ * 1.60kg * v²

solves to v = 1.03 m/s

b) ½ * 990N/m * (0.0470m)² = ½ * 1.60kg * v²

v = 1.17 m/s

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