# Homework Help: Find time given a force with respect to displacement

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1. Jun 24, 2017

### Kiwigami

1. The problem statement, all variables and given/known data
With a function that gives Force with respect to Displacement (F(x) = 2000-100x), is there a way to find a general formula for the time it takes from displacement 0 to 20? The force is acting on an object, pushing it horizontally. Assume no friction and no air resistance.

2. Relevant equations
Force with respect to Displacement (x):
F(x) = 2000-100x
At 0 displacement, there's 2000 Newton of force acting on the object, and the force linearly decreases until there's no force at 20 meters away and beyond.

3. The attempt at a solution

Work = ∫ from 0 to b of (F * dx) = ∫ from 0 to b of (2000-100x) dx = -50(b- 40)b
Work = Final Kinetic Energy - Initial Kinetic Energy. Initially, the object is stationary with no initial velocity, so I'm assuming that the initial Kinetic Energy is 0.

Work(b) = Kinetic Energy(b) = -50(b - 40)b = 0.5mv^2.

Solving for v, I get: v(b) = sqrt((2/m) * -50(b - 40)b)

From here, I'm tempted to plug velocity into this formula (Displacement = Velocity * Time) and solve for time, but I feel like this won't work.

2. Jun 24, 2017

### olgerm

I think you only need Newtons II law to solve this. You get 2. order differencial equation. General formula has mass as variable.

3. Jun 25, 2017

### haruspex

That is only true for constant velocity. The general expression is the integral of velocity wrt time. That will give you a differential equation to solve.
The first integration stage from that equation produces the energy relation Kiwigami obtained, so there is no benefit in going that route.

4. Jun 25, 2017

### olgerm

I meant integrating over time.

5. Jun 25, 2017

### haruspex

How are you going to do that, given acceleration as a function of distance?

6. Jun 25, 2017

### olgerm

just write Newton's II law($\frac{\partial^2 x}{\partial t^2}=F/m$) and replace F to function, that he gave. Doing that will result getting second-order linear ordinary differential equation. Also is known that displacement and speed at time 0 are 0.

7. Jun 25, 2017

### haruspex

You did not answer my question. How are you going to integrate F, a function of x, with respect to time?

8. Jun 25, 2017

### olgerm

By writing Newton's II law($\frac{\partial^2 x}{\partial t^2}=F/m$) and replacing force F to function, that he gave and taking undefined time-integral of both sides. Taking this integral is not necessary.
$\frac{\partial^2 x}{\partial t^2}=\frac{2000-100x}{m}$ ⇔ $x+k_1+k_2 \cdot t=\iint(dt^2 \cdot \frac{2000-100x}{m})$ .
by $x(0)=0$ and $\frac{\partial x}{\partial t}(0)=0$ we know that $k_1=0$ and $k_2=0$.

Last edited: Jun 25, 2017
9. Jun 25, 2017

### haruspex

I ask a third time, how are you going to perform that integral of x wrt time? Writing down the integral doesn't answer it, you actually need to integrate it.

10. Jun 25, 2017

### olgerm

It ($\frac{\partial^2 x}{\partial t^2}=\frac{2000-100x}{m}$) is called differential equation. Before solving it I cant find that $\iint(dt^2 \cdot \frac{2000-100x}{m})+k_3+k_4 \cdot t= c_2 \cdot sin(\frac{10 \cdot t}{\sqrt{m}}) + c_1 \cdot cos(\frac{10\cdot t}{\sqrt{m}}) +20$
, but after solving it I can.

11. Jun 25, 2017

### haruspex

But having solved the D.E. by some other method, you do not need to do this integral, so writing it out as a double integral of x wrt t was a complete waste of time. Including mine.

12. Jun 25, 2017

### TSny

What if you repeat the above calculation, but you integrate from 0 to some arbitrary value of x between 0 and b? Would you then be able to obtain an expression for the velocity as a function of x?

13. Jun 25, 2017

### Ray Vickson

Things simplify if you recognize that you can write $F = -100(x-20)$, so in terms of the new displacement variable $y = x - 20$ the force law is just $F = - 100y$. This is Hooke's law (for stretching forces in a spring), so a Google search on "Hookes law" might be very helpful.

Last edited: Jun 25, 2017