Answer:
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
The magnitude of the acceleration of the block is [tex]30 \;\rm m/s^{2}[/tex] 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,
[tex]a =- \omega^{2} \times x[/tex] (Negative sign shows that the acceleration is opposite to displacement)
Here, [tex]\omega[/tex] is the angular frequency of oscillation. And its value is,
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
Then solving as,
[tex]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}[/tex]
Now, calculate the force to obtain its direction as well.
[tex]F = ma \\\\F = 0.5 \times (-30)\\\\F = -15 \;\rm N[/tex]
Clearly, force is directed along negative direction of x -axis.
Thus, we can conclude that the magnitude of the acceleration of the block is [tex]30 \;\rm m/s^{2}[/tex] and the direction of force is toward negative x -axis.
Learn more about the simple harmonic motion here:
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