Find time given a force with respect to displacement

[Kiwigami]

Homework Statement

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.

Homework 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.

The Attempt at a Solution

[/B]
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.

 

[olgerm]

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

 

[haruspex]

Displacement = Velocity * Time

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.

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

The first integration stage from that equation produces the energy relation Kiwigami obtained, so there is no benefit in going that route.

 

[olgerm]

The first integration stage from that equation produces the energy relation Kiwigami obtained, so there is no benefit in going that route.

I meant integrating over time.

 

[haruspex]

I meant integrating over time.

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

 

[olgerm]

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

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.

 

[haruspex]

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.

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

 

[olgerm]

How are you going to integrate F, a function of x, with respect to time?

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 $k_2=0$.

 

[haruspex]

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 $k_2=0$.

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.

 

[olgerm]

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.

It ($\frac{\partial^2 x}{\partial t^2}=\frac{2000-100x}{m}$) is called differential equation. Before solving it I can't 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.

 

[haruspex]

It ($\frac{\partial^2 x}{\partial t^2}=\frac{2000-100x}{m}$) is called differential equation. Before solving it I can't 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.

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.

 

[TSny]

The Attempt at a Solution

[/B]
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)

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?

 

[Ray Vickson]

Homework Statement

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.

Homework 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.

The Attempt at a Solution

[/B]
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.

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.