A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?
2 answers:
The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.
<h3>What is spring force?</h3>The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is
F = kx
The given data in the problem is;
F is the spring force =200
K is the spring constant= 2500 N/m
x is the length by which spring got stretched =?
The stretch of the spring is found as;
Hence the length by which the spring got stretched will be 0.08 m.
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The acceleration of the runner in the given time is 2.06m/s².
Given the data in the question;
Since the runner begins from rest,
- Initial velocity;
- Final velocity;
- Time elapsed;
Acceleration of the runner;
Velocity is the speed at which an object moves in a particular direction.
Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
To determine the acceleration of the runner, we substitute our given values into the equation above.
Therefore, the acceleration of the runner in the given time is 2.06m/s².
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