A child's toy consists of a piece of plastic attached to a spring, as shown below. The spring is compressed against the floor a distance 1.8 cm and released. If the spring constant is 107 N/m, what is the magnitude of the spring force acting on the toy at the moment it is released?
N
(0.018m)(107N/m)=1.926N
To find the magnitude of the spring force acting on the toy at the moment it is released, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is: F = -kx
Where:
F is the spring force
k is the spring constant
x is the displacement from the equilibrium position
In this case, the displacement of the spring is given as 1.8 cm (which is 0.018 m) and the spring constant is given as 107 N/m.
Substituting these values into the formula, we get:
F = -(107 N/m)(0.018 m)
Calculating this, we find:
F ≈ -1.926 N
However, since we are looking for the magnitude of the spring force (which is always positive), we can take the absolute value of the force, resulting in:
F ≈ 1.926 N
Therefore, the magnitude of the spring force acting on the toy at the moment it is released is approximately 1.926 N.