Question

A person exerts a 19-N force on a cart attached to a spring and holds the cart steady. The cart is displaced 0.060 m from its equilibrium position. When the person stops holding the cart, the system cart spring undergoes simple harmonic motion. Determine the spring constant of the spring. Express your answer with the appropriate units.

Answer 1

Answer:

 K = 316.66 N/m

Explanation:

Given that

F= 19 N

Displacement .x= 0.06 m

We know that spring force F given as

$F=k x$

F = force

K = Spring constant

x = Displacement

Now by putting the values in the above equation

$19 = 0.06\times K$

$K=\dfrac{19}{0.06}$

K = 316.66 N/m

Therefore the value of spring constant will be 316.66 N/m.