Finding the force of reaction as a function of time

[jumboopizza]

Homework Statement

the only part of the question I am unsure about is finding the force of reaction as a function of time

Homework Equations

F=Ma
F=-Kx

The Attempt at a Solution

F=Ma
F=M(Dv/Dt)

F Dt= M Dv

i integrate force with respect to time from final time=t and initial time = 0

integrating M with respect to velocity from final velocity=V and initial velocity=0

I get F(t)=MV

is this the correct way to get the force as a function of time?

 

[na na]

did you find the solution

 

[PeroK]

did you find the solution

This thread is ten years old!

 

[na na]

I know but I also have a problem that is very similar to that one . how do you find the force as a function of time

 

[Delta2]

Make a new thread in homework forums please and be sure to post as many details as you can about your problem and attempted solution, including the exact statement of the problem as it is given to you.

What the OP does here is not entirely correct. From $Fdt=MdV$ it is not correct to conclude $F(t)=MV$, it should have been $dF=MdV$ in order to conclude that.

 

[na na]

Make a new thread in homework forums please and be sure to post as many details as you can about your problem and attempted solution, including the exact statement of the problem as it is given to you.

What the OP does here is not entirely correct. From $Fdt=MdV$ it is not correct to conclude $F(t)=MV$, it should have been $dF=MdV$ in order to conclude that.

ok thank you very much

 

[jbriggs444]

Make a new thread in homework forums please and be sure to post as many details as you can about your problem and attempted solution, including the exact statement of the problem as it is given to you.

What the OP does here is not entirely correct. From $Fdt=MdV$ it is not correct to conclude $F(t)=MV$, it should have been $dF=MdV$ in order to conclude that.

Ummm, make that $F = M \frac{dV}{dt}$. Which would more conventionally have used a lower case M and V and would be commonly written as: $F=ma$

One could start with $F=\frac{dp}{dt}$, take the definition of momentum $p$ as $mv$, substitute to arrive at $F=\frac{d(mv)}{dt}$ use the product rule for differentiation and get $F=m \frac{dv}{dt} + v \frac{dm}{dt}$, take m as a constant so that its derivative is zero to get $F = m \frac{dv}{dt}$ and then use the definition of acceleration to rewrite that as $F=ma$.