A 2.0 kg mass is fastened to a spring with force constant of 45 N/m. The mass is held motionless by a student at a position 15 cm below the spring's natural unstretched position. When the ball is released, what is the Fnet acting on it? What is the acceleration of the ball at that point?
F = .15 * 45
a = F/m
The first question asks for Fnet, not the force applied.
The force applied/stored is: Fs = kΔx
The net force is: Fnet = Fg - Fs (Assume down is +ve)
From this we get a positive force (Fg is greater than Fs), i.e. when the ball is released, it accelerates downwards. (The spring does not have enough force to pull the ball back.)
The acceleration is positive and can be obtained from: Fnet = ma
Can anyone confirm?
a=3.375m/s^2
To determine the net force acting on the mass when it is released, first, we need to calculate the force exerted by the spring at that position. We know that the force exerted by a spring can be given by Hooke's Law: F = -kx, where F is the force, k is the force constant, and x is the displacement from the equilibrium position.
In this case, the mass is held 15 cm (or 0.15 m) below the unstretched position, so the displacement from the equilibrium position is -0.15 m (negative because it's below).
Using Hooke's Law, we can calculate the force exerted by the spring:
F_spring = -kx = -(45 N/m)(-0.15 m) = 6.75 N
Since the mass is initially held motionless, the net force acting on it is zero, as it balanced the force exerted by the spring:
F_net = 0 N
The acceleration of an object can be calculated using Newton's second law: F = ma, where F is the net force acting on the mass, m is the mass, and a is the acceleration.
In this case, since the net force is zero, the acceleration of the mass at that point is also zero:
a = 0 m/s^2