F = .15 * 45
a = F/m
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?
3 answers
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?
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
1 answer