Suppose a block of mass, m is connected to a spring of force constant, k. The block is moved on a frictionless surface through distance, b with the help of a uniform external force, F so that the spring gets stretched.
What will be the velocity of the block when it completes the distance, b?

Known Quantities:-
Mass - m
Force Constant - k
Distance - b
Force - F

Unknown Quantity: Velocity - v

Note: The mass of the spring and the velocity with which it is moved with the block is considered negligible in both the following methods.

Method 1:-

When the block is moved(consequently, the spring will stretch), the work done by the external force will get stored as Potential Energy(PE) in the spring and also, will appear as Kinetic Energy(KE) of the block.

Initally, the block is at rest, so $KE_i=0$ and at equilibrium position, so $PE_i=0$

$W=\vec{F}\cdot \vec{b} = F b cos0 $

$W= \Delta PE(Spring) + \Delta KE(Block)$
$W= (PE_f - PE_i) + (KE_f - KE_i)$

$Fb= \frac{1}{2}kb^2 + \frac{1}{2}mv^2$

$v^2= \frac{2Fb-kb^2}{m}$


My book solved the problem with this method.
But I tried a different method where I used Newton's equations of motion.

Method 2:-

As the Force is uniform, so the acceleration, a produced in the block will also be uniform. So, we can apply Newton's Equations Of Motion.



Initially, the block was at rest, so intital velocity, $u=o$.

Applying 3rd equation of motion,



Now, I just want to make sure that is the method correct or not.
If it's correct then, both the methods 1 and 2 should give same velocity.

But, that's the real problem that the values of velocity from method 1 and method 2 are not matching up.

Please, figure me out.

  • $\begingroup$ With simple harmonic motion, $a \propto -x$ which means that the acceleration is not constant. Just because there is a constant force F on the block doesn't mean the net force will be constant. $\endgroup$ – Jordan Abbott Nov 20 '17 at 17:24
  • $\begingroup$ The net force acting on the block is not constant, because the spring provides a restorative force $F=kx$ that opposes the external (constant) force. $\endgroup$ – Gert Nov 20 '17 at 17:27
  • $\begingroup$ Ok, I don't how this question is off-topic. I didn't put the tag homework-and-excercises when I posted this question. I ask about whether I can solve it using Equations of Motion or not(which I think comes under the Physics domain) and though I have shown 2 different methods of solving it, the reason for putting it on hold is "show some effort" and "ask about a specific physics concept". Can anyone explain this??? $\endgroup$ – lakhi Nov 26 '17 at 7:28

You got stuck with the second method because you're using an incorrect Newton's Equation of Motion. The correct one is (the object travels on the $x$ axis), where $F$ is a constant force acting on the block:

$$ma=F-kx$$ $$m\ddot{x}+kx-F=0$$

But it's very clear that the method of conservation of energy is far simpler. Have a look at this derivation.


$$kx-F=X$$ So: $$k\dot{x}=\dot{X}$$ $$k\ddot{x}=\ddot{X}\implies \ddot{x}=\frac1k\ddot{X}$$ $$\frac{m}{k}\ddot{X}+X=0$$ $$\ddot{X}+\frac kmX=0$$ So the differential equation is now homogeneous:


$$\frac km=\omega^2$$

$$\ddot{X}+\omega^2 X=0$$ This classic second order differential equation solves to:

$$X(t)=A\cos(\omega t+\phi)$$ Where $A$ and $\phi$ are constants (amplitude and phase angle, resp.) that we determine from the initial conditions. So:

Assume at $t=0$, $x=0 \implies X(0)=-F$

$$-F=A\cos\phi$$ Assume at $t=0, \dot{x}=0 \implies \dot{X}(0)=0$.

$$\dot{X}=-A\omega\sin(\omega t+\phi)$$ $$\dot{X}(0)=-A\omega\sin\phi=0\implies \phi=0$$ $$A=-F$$ $$X(t)=-F\cos\omega t=kx(t)-F$$ $$\boxed{x(t)=\frac Fk(1-\cos\omega t)}$$ So this is the equation of motion. And this is the expression for velocity in time: $$\boxed{\dot{x}=\frac Fk \omega \sin \omega t}$$

Use these to find the time needed to travel the distance $b$, then use that time, say $t_b$, to determine $\dot{x}(t_b)$ with the velocity expression.

So that's quite a kerfuffle, compared to the conservation method, which explains why your textbook recommends it. The object lesson is that if you're not tasked to find the exact trajectory of the object(s), solve this kind of problem with conservation, preferably.

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.