A spring has a spring stiffness constant k=9.5 N/m, how much must this spring be stretched to store 189.7 J of potential energy by applying a constant force of 1253.1?
Energy stored = (1/2) k x^2
189.7 = (1/2) (9.5) x^2
x^2 = 39.9
x = 6.32 meters
now can you do that with 1253.1 N ?
1253.1 = 8.5 x
x = 147 meters
so yes
is it 8.5 or 9.5? thank you btw
To determine how much the spring must be stretched, we can use Hooke's Law, which relates the force applied to a spring to the amount the spring is stretched. Hooke's Law is given by the equation:
F = k * x
Where:
F is the force applied to the spring,
k is the spring stiffness constant (also known as the spring constant),
x is the displacement or stretch in the spring.
In this case, we are given the spring stiffness constant k = 9.5 N/m and the force applied F = 1253.1 N. We need to find the displacement x.
Rearranging the equation, we have:
x = F / k
Plugging in the values we have:
x = 1253.1 N / 9.5 N/m
Now we can calculate the value of x:
x = 132.06 m
Therefore, the spring must be stretched by approximately 132.06 meters to store 189.7 J of potential energy by applying a constant force of 1253.1 N.