Physics, 30.03.2020 17:28 Meliiiii

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 ?

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The magnitude of the acceleration of the block is $30 \;\rm m/s^{2}$ 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,

$a =- \omega^{2} \times x$ (Negative sign shows that the acceleration is opposite to displacement)

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

$\omega = \sqrt{\dfrac{k}{m}}$

Then solving as,

$a = - (\sqrt{k/m} )^{2} \times x\\\\a = - \dfrac{k}{m} \times x\\\\a =- \dfrac{50}{0.5} \times 0.3\\\\a = -30 \;\rm m/s^{2}$

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

$F = ma \\\\F = 0.5 \times (-30)\\\\F = -15 \;\rm N$

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

Thus, we can conclude that the magnitude of the acceleration of the block is $30 \;\rm m/s^{2}$ and the direction of force is toward negative x -axis.

Learn more about the simple harmonic motion here:

link

Answer from: juliana0122

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

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