Theory: the stretching force as long as the elastic limit of the spring is not e

Question

Theory: the stretching force as long as the elastic limit of the spring is not exceeded. Thus a force F produces an elongation or compression, x, in the spring, F, kx, where k is the force constant of the spring. The spring stores energy in the form of potential energy, U, where U =-kr'. The total energy of the systemis given by E = U + K =-kx 2 + 1 mw2. As energy is always conserved the maximum kinetic and potential energies are such that I,na,--kx? = Kmax-, m v2 (where x and v are the maximum extension of the spring and maximum velocity respectively). According to Hooke's law, the elongation of a stretched spring is proportional to Question 1: How does the energy change if the spring is stretched from its initial extension of 2 cm to an extension of 4 cm? What happens to the force in the spring if this initial distance is halved to 1

Explanation / Answer

# Initial length = 2 cm.

Final length = 4 cm.

Since the spring is stretched and held at 4 cm extension it will contain potential energy only at initial and finaL position.

so its energy will change as

E(initial)= 1/2 ×k× (2)^2= 2k.

E(final) = 1/2×k×(4)^2 = 8k

Energy will INCREASE by a factor of 4. It means energy become 4 times than initial energy.

#Let initial distance is d.

So force = kd

Now when halved it become d/2.

Now force= kd/2

So force Decreases by a factor of 2. It means force is also HALVED.

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