Block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . The other end of the spring is attached to a wall, and there is negligible friction between the block and the horizontal surface. When the spring is unstretched, the block is located at x=0m . The block is then pulled to x=0.3m and released from rest so that the block-spring system oscillates between x=−0.3m and x=0.3m . What is the magnitude of the acceleration of the block and the direction of the net force exerted on the block when it is located at x=0.3m ?

The magnitude of the acceleration of the block is   and the direction of force is toward negative x -axis.

Given data:

The mass of block is, m = 0.5 kg.

The value of spring constant is, k = 50 N/m.

The stretching distance is, x = 0.3 m.

To solve this problem, the concept of simple harmonic motion can be applied. Where, the oscillation of spring- mass system is determined with amplitude. And in this problem, the stretching distance is the amplitude.

So, the Acceleration is given as,

           (Negative sign shows that the acceleration is opposite to displacement)

Here, is the angular frequency of oscillation. And its value is,

Then solving as,

Now, calculate the force to obtain its direction as well.

Clearly, force is directed along negative direction of x -axis.

Thus, we can conclude that the magnitude of the acceleration of the block is   and the direction of force is toward negative x -axis.

Learn more about the simple harmonic motion here:

link

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

es el número de veces que aparece o sucede una cosa durante un periodo.