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Schach [20]
3 years ago
6

if a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of the spring?

Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

0.1 N

Explanation:

Considering the relationship between force,

spring constant and extension as defined by Hook's law

The force F=xk as from Hooke's law where F is the force of the spring, k is spring constant and x is extension or compression. Substituting 2 N/m for k and 5cm which is equivalent to 0.05 m for extention x then the force will be

F=2*0.05=0.1 N

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weqwewe [10]

The magnitude of force acting on the bumper is 3760 N.

<h3>What is Work energy theorem?</h3>

It states that the Work done in moving a body is equal to the change in kinetic energy of the body

Kinetic energy = 1/2 mv²

Given is a car's bumper designed to withstand 4.32 km/h or 1.2 m/s collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance.

The cushion collapses 0.180 m while bringing 940 kg car to rest from a initial speed of 1.2 m/s

Work done = Force x displacement

As the displacement of the bumper and force acted on it is in same direction, so the work done is

W = Fxcos0° = Fx

The body is coming to rest, so, final velocity is zero. Then, change in kinetic energy will be

ΔK.E = K.Ef - K.Ei

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W = Fx  = m/2 (v² - u²)

Substitute the value and calculate the force,

F  = [940 x (0² - 1.2²)] / 2x0.180

F = 3760 N

Thus, the magnitude of force is 3760 N.

Learn more about work energy theorem.

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