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gashan · Sep 25, 2010, 11:54 AM

A body of mass M is moving along a rough horizontal surface with velocity v. The body is being acted upon by a force 4M newtons in the direction of motion and a resistive force of magnitude kv. The displacement of the body from the origin at any time is denoted by x.
I) Form the differential equation which describes the motion of the body in terms of M,x,k and t.
ii)if M=6 and k=0.5. determine the general solution of the differential equation derived in (I) above.

galactus · Sep 25, 2010, 02:39 PM

You have the force acting upon the body and the resistive force.

So, since F=ma, we have:

$M\cdot \frac{d^{2}x}{dt^{2}}=4Mg-kv$

g=gravity constant of 9.81 m/s^2.

My physics is rusty. Perhaps someone can confirm or deny my DE?

gashan · Sep 25, 2010, 02:43 PM

Yes I can. Thank you

gashan · Sep 25, 2010, 03:05 PM

you are making reference to the actual solving of the equation with regards to substitution, right?

gashan · Sep 25, 2010, 03:09 PM

the initial formula is correct from a physics stand point. I was trying to assist with this question(as I am helping a friend), and I got a similar equation

galactus · Sep 25, 2010, 03:59 PM

To solve the DE

$x=\frac{981t^{2}}{50}-\frac{kvt^{2}}{2M}+C_{1}t+C_{2}$

Unknown008 · Sep 25, 2010, 11:08 PM

I tried yesterday, but got something I didn't know how to continue.

Here's what I would have posted:

The displacement of the body is given by x.

But x itself depends on v;

$\frac{dx}{dt} = v$

Now, the current acceleration of the body is given by Newton's equation:

F = ma
4M - kv = Ma
a = 4 - (kv)/M

And

$\frac{dv}{dt} = a$

So:

$v = \int a\ dt$

$v = \int 4 - \frac{kv}{M}\ dt$

$\frac{dx}{dt} = \int 4 - \frac{kv}{M}\ dt$

But then, I don't know how to remove the v in the integral, as in the question, express in terms of x, k, M and t only.

In your equation galactus, you don't have 4Mg because that is not the forward force.

Gravity is not involved in here as its line of action is perpendicular to the forces involved. If friction was involved, then gravity would too be involved. But the question only stated resistive forces, so, gravity is not involved.

I'll have to think more about this.

Unknown008 · Sep 25, 2010, 11:12 PM

Ooh, I think I know...

$\frac{dx}{dt} = \int 4\ dt - \frac{k}{M} \int v\ dt$

We know that the integral of v with respect to time gives the displacement, so, we get:

$\frac{dx}{dt} = 4t - \frac{k}{M}(x)$

$\frac{dx}{dt} = 4t - \frac{kx}{M}$