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ololo11 [35]
3 years ago
6

A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s

imple harmonic motion of amplitude A. m k −A 0 +A vx0 Which statement about the block is correct?a. At x = A, its acceleration is zero.b. At x = 0, its velocity is zero.c. At x = A, its velocity is at a maximum.d. At x = 0, its acceleration is at a maxi
Physics
1 answer:
AleksandrR [38]3 years ago
8 0

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

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Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

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s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

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3 0
3 years ago
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