The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
1 answer:
Answer:
Force constant will be 1195.85 N/m
Work done will be 1.6859 J
Explanation:
We have given the force, F = 63.5 N
Spring is stretched by 5.31 cm
So x = 0.0531 m
Force is given , F = 63.5 N
We know that force is given by
So
k = 1195.85 N/m
Now we have to find the work done
We know that work done is given by
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1) Maximum
2) Maximum
Explanation:
The force acting on a mass on a spring is given by Hooke's law; in magnitude:
where
F is the force
k is the spring constant
x is the displacement
Also we know from Newton's second law that we can write
where
m is the mass
a is the acceleration
So we can write the equation as
(1)
From this relationship, we see that the acceleration is directly proportional to the displacement.
On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by
(2)
where the first term is the elastic potential energy while the second term is the kinetic energy, and where
v is the velocity of the mass
From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.
Therefore this means that:
- When acceleration increases, velocity decreases
- When acceleration decreases, velocity increases
Therefore, the two answers here are:
- When the acceleration of a mass on a spring is zero, the velocity is at a maximum
When the velocity of a mass on a spring is zero, the acceleration is at a maximum