friction problem

Question

As long as you don’t stretch it too much, the force exerted by the spring obeys Hooke’s Law. The further you stretch the spring, the harder it pulls back on you. The stiffness of a spring is represented by the spring constant (see Section 16.1 in the text). In this problem consider the spring to be of negligible mass.

Suppose you stretch a spring 0.0444 m from its equilibrium position. The spring has a spring constant of 1.37×10³ N/m. What is the magnitude of the force you need to exert on the spring?

Suppose you place this same spring on a hook and hang a 0.631 kg object on the end. How far does the spring stretch from equilibrium? (think about the force that the object exerts on the spring).

Answer 1

1) There's only one force on the spring considering the spring isn't vertical. F_spring = kx, so plug in the spring constant and the distance stretched and solve for the force.

2) When it's vertical, we have two forces. We have the weight of the spring and the object pulling downward and the force of the spring pulling upward against it.

Use the formula, ∑F_y = ma_y. This says that the sum of the forces in the y-direction (vertical direction) equals the mass times the acceleration in the vertical direction. The object is just hanging there so we have no acceleration up or down. So a_y = 0. Do you think that you can solve for the distance the spring was stretched?

The formula is reduced to ∑F_y = 0.