I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

Now, determine the amplitude of oscillation, A;

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

Therefore, the amplitude of the oscillation is 2.82 cm

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.