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aliya0001 [1]
3 years ago
11

A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

you stretch the spring to a total length of 27 cm (0.27 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 11 cm (0.11 m)?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

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Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang
Marrrta [24]

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2/R*sqrt (g*s*sin(θ)) = w

Explanation:

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- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

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- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

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We have:

                              m*g*s*sin(θ) = 0.25*m*R^2*w^2

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3 years ago
Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
nirvana33 [79]

Answer:

1.327363 m/s

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Explanation:

u = Initial velocity

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Energy

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The flea will travel 0.00090243 m upward

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3 years ago
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