The elastic potential energy of the spring is 6.8 J
Explanation:
The elastic potential energy of a compressed/stretched spring is given by the equation:
where
k is the spring constant
x is the elongation of the spring
The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):
By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,
Therefore when the spring has a elongation of , its potential energy is
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Since sphere is performing pure rolling motion
So here the contact point of sphere with respect to inclined plane is always at rest.
This contact point will not move at any instant and and hence sphere will perform pure rolling.
So here we can say that sphere will have some friction force at the contact point but there is no displacement of that point.
So the work done by this static friction will always ZERO.
So there is no work done by friction on the sphere while it is in pure rolling.
Answer:
The position function is .
Explanation:
Given that,
Acceleration
Initial velocity
Initial displacement
We know that,
The acceleration is the rate of change of velocity of the particle.
The velocity is the rate of change of position of the particle
We need to calculate the the position
The acceleration is
On integration both side
At t = 0
Now, On integration again both side
At t = 0
Hence, The position function is .