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MakcuM [25]
3 years ago
7

A spring with spring-constant of 100 n/m is compressed 0.10 m. what is its maximum stored elastic potential energy

Physics
1 answer:
Ganezh [65]3 years ago
5 0

Elastic potential energy stored in a spring is

(1/2) · (spring constant) · (stretch or compress)² .

PE = (1/2) · (100 N/m) · (0.1 m)²

PE = (50 N/m) · (0.01 m²)

PE = (50 · 0.01) (N · m / m²)

PE = 0.5 N · m

PE = 0.5 Joule

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3 years ago
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5 0
3 years ago
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Answer:

The velocity of the launcher after the projectile is launched is  -5.011 m/s

Explanation:

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Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

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Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} =  \frac{-461}{92} m/s

v₃ = -5.011 m/s.

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