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Arisa [49]
3 years ago
15

A spring is laid horizontally on a table and a force of 62 N is required to stretch the spring 12 cm. The spring is then held ve

rtically and an object is attached to the end of it which causes the spring to stretch 9.7 cm. What is the Mass of the object?
SHOW YOUR WORK
Physics
1 answer:
Talja [164]3 years ago
8 0

Answer:

5.11 kg

Explanation:

Hooke's law states that stress is directly proportional to strain. It can be represented by the equation:

F = -kx

Where x is the displacement of the spring’s end from its equilibrium position, F is the force applied to the spring, k is a constant known as spring constant.

At first a force (F) of 62 N i used to stretch a spring by 12 cm (x).

Substituting into Hooke's equation:

F = kx

k = F/x = 62 / 12 = 5.17 N/cm

The spring is then held vertically and an object. The object is acted by upon by acceleration due to gravity since it is vertically upward. Therefore the force created by the object F = mg where g = acceleration due to gravity = 9.8 m/s². It causes a stretch (x) of 9.7 cm

Therefore:

mg =  5.17 N/cm × 9.7 cm

9.81m = 50.117

m = 50.117 / 9.81 = 5.11 kg

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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

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The maximum force F

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3 0
3 years ago
U235 + n → Xe134 + Sr100 + 2n
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Answer:

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Hope it helps and have a wonderful day!

4 0
3 years ago
Read 2 more answers
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