The force exerted by the spring on the sphere can be calculated using Hooke's Law:
F = kΔx
Where F is the force, k is the spring constant, and Δx is the change in length of the spring.
In this case, the change in length is given as:
Δx = 0.0950 m - 0.100 m = -0.005 m
Since the spring is compressed in this case, the force will act in the opposite direction. Therefore, the negative sign is added to the equation.
F = -kΔx
F = -(450 N/m)(-0.005 m)
F = 2.25 N
The magnitude of the force exerted on the sphere is 2.25 N.
The direction of the force is upward since the spring is compressed and wants to return to its relaxed length.
Therefore, the electric force exerted on the sphere is 2.25 N upward.
A plastic spring with spring constant 450 N/m has a relaxed length of 0.100 m . The spring is positioned vertically on a table, and a charged plastic 1.20-kg sphere is placed on the top end of the spring. Another charged object is suspended above the sphere without making contact. If the length of the spring is now 0.0950 m , what are the magnitude and direction of the electric force exerted on the sphere?
1 answer
1 answer