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Amanda [17]
3 years ago
5

one end of a horizontal spring(k=80N/m)is held fixed while an external force is applied to the free end,stretching it slowly fro

m×1=o to ×2=4cm.what is work done by the applied force on the spring​
Physics
1 answer:
loris [4]3 years ago
6 0

Answer:

6.4 Joules

Explanation:

For springs, Hooke's law states that;

F = ke

where F is a force applied, e is the extension and k is the spring constant.

Work done in a spring is the same as the potential energy stored in the spring. So that;

Work done = \frac{1}{2} ke^{2}

e = x_{2} - x_{1}

x_{2} = 4 cm = 0.4 m

x_{1} = 0

So that,

x_{2} - x_{1} = 0.4 m

Thus,

Work done = \frac{1}{2} x 80 x (0.4)^{2}

                   = 40 x 0.16

                   = 6.4

Work done = 6.4 J

The work done by the force on the spring is 6.4 Joules.

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4 years ago
Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For
SpyIntel [72]

Answer:

1.27\times 10^{12}\Omega/m

Explanation:

We are given that

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A force of 42N is needed to start a box sliding across the floor. The weight of the box is 55N.
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