a block of mass 0.1kg is attached to spring and placed on a horizontal frictionless table .the spring is stretched 20cm when a force of 5n is applied.calculate the spring constant
F=kx
K=5/0.2=50/2=25N/M
T=2π✓m/K
T=2×3.142✓0.1/25
T=6.284×0.0632
T=0.397Sec
Yes
To calculate the spring constant, we need to use Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = -k * x
Where:
F is the force applied to the spring,
k is the spring constant,
x is the displacement of the spring.
In this case, we are given that a force of 5N is applied and the spring is stretched by 20cm (or 0.2m). We can substitute these values into the equation to solve for the spring constant.
5N = - k * 0.2m
First, let's isolate the spring constant (k):
k = -5N / 0.2m
We can calculate the spring constant by dividing the force by the displacement:
k = -5N / 0.2m
k = -25 N/m
Therefore, the spring constant (k) is -25 N/m.