# Velocity is zero, but acceleration is not?

Imagine a block (of mass m) attached to a spring (of spring constant k) that is hanging from a fixed support on the ceiling. The spring is initially in its relaxed state( no compression or extension). At this moment velocity of the block is zero

When I release the spring from rest, the force due to gravity and the force due to spring(restoring force) are the two forces acting on the block.

According to me,when:

magnitude of Force due to gravity> magnitude of Restoring force, then the velocity of the body increases.

magnitude of Force due to gravity=magnitude of Restoring force, then the acceleration of the body is zero, and the body has maximum constant velocity.

magnitude of Force due to gravity < magnitude of Force of the spring, then the velocity of the body decreases until it stops.

However when the spring reaches maximum extension, there is still negative accelaration, but the velocity of the body is zero.

[ I used work-energy theorem to calculate maximum extension as $\frac{2mg}{k}$, and therefore, force due to spring = 2mg,

force due to gravity= mg,

and net force = force due to spring- force do to gravity=mg]

If there is a net upward force, then there is also an acceleration, but the velocity is zero. How is this possible?

• Are you troubled that an object can have a zero velocity and at the same time a nonzero acceleration? Isn't this necessary for any stationary object to start moving? – M. Enns Oct 22 '17 at 13:50
• Because a = dv/dt. V soon gets different than zero. Perhaps you are not into calculus... Just consider the acceleration as that around a very small interval centered on the point (elongation) for which V = 0 – Alchimista Oct 22 '17 at 13:50
• Would you say that the spring is undergoing harmonic motion? – GreenApple Oct 22 '17 at 14:21
• It’s a bad idea to consider only magnitudes since force is a vector. In this specific problem it is even worse as the force from the spring is the same for same compression and extension, but has different direction. You might want to redo your analysis considering the directions of the forces. – ZeroTheHero Oct 22 '17 at 16:48

Try to compare it with this: you thrown an object vertically upwards with a velocity $v_{0}$, so the only force acting on the object is gravity downwards with an acceleration $-g$. At a certain point, the object will slow down to a velocity of $0$ and then fall back downwards to earth. In this whole process, the acceleration is constant, but the velocity is still zero at its highest point.
More mathematically: $a = -g = \frac{dv}{dt}$. This is a simple differential equation and can be solved easily with integrals: $-gdt = dv$ so $\int_{t_0}^{t}-gdt = -g\int_{t_0}^{t}dt = -g\cdot(t-t_0) = -g\cdot t$ if we take $t_0 = 0s$ and also $\int_{v_0}^vdv = v(t)-v_0$, and so we get $-g\cdot t = v(t)-v_0$ or $v(t) = v_0 - g\cdot t$. Acceleration $a$ is constant and thus $a(t) = -g$. Now $v(t)$ will be equal to zero for $t = \frac{v_0}{g}$, so $v(\frac{v_0}{g}) = 0$ but also $a(\frac{v_0}{g}) = -g \neq 0$.
• I'm not sure what you mean with time = infinity. Do you mean that after a while, the object will stop moving and go to rest (at velocity 0)? This is the case if you have damping (=friction, a force in the opposite direction of the spring force, usually with respect to velocity, $F_d = b\cdot v$) where the system will "lose" a bit of energy during the motion and eventually go to a position where $k\cdot \Delta x = m\cdot g$, with no velocity or net force. In an ideal case, there is no damping and the object will oscillate forever (with zero velocity only at maximum extension). – BMike Oct 23 '17 at 21:42
• As for the last part: in a harmonic motion, the acceleration and velocity will indeed change sinusoidally with respect to time. Only, they are out of phase, meaning that velocity will be max when acceleration is 0 and when velocity is zero, acceleration will be max. Look at the following: $v(t) = \omega A \sin(\omega\cdot t)$. Since $a(t) = \frac{dv}{dt}$, we have $a(t) = \omega^2 A \cos(\omega \cdot t) = \omega^2 A \sin(\omega \cdot t + \frac{\pi}{2})$. As you can see, both are sinusoidally, but they don't "start" sinusoidally at the same time. – BMike Oct 23 '17 at 21:53