Newton's law question involving force and acceleration

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The discussion revolves around a physics problem involving Newton's laws, specifically the relationship between force, mass, and acceleration. The force acting on a mass is given by F = b + ct^2, and participants emphasize the need to apply Newton's second law, F = m(dv/dt), to find the velocity as a function of time. A correct approach involves equating the given force to the expression from Newton's law and solving for velocity. The initial response indicates that the final expression for velocity is V = V0 + bt + (ct^3)/3. Participants are reminded to demonstrate their understanding and efforts before receiving further assistance.
Arsene11
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New poster has been reminded to always show their work when starting schoolwork threads
Homework Statement
A body of mass m = 1 kg moves along a straight line with speed v0, when it is
the action of a force in the same direction of motion. The magnitude of this force is given in terms of
of time t in seconds by the following expression:
F = b + ct^2

where b and which are constants with appropriate relationships for the term-owning unit of
force in newtons. Find an expression for the velocity v of the body as a function of time t.
Relevant Equations
F = b + ct^2
I have no idea what to do but the answer is V=V0+ bt +(ct^3)/3
 
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Hi Arsene11 and welcome to PF.

:welcome:
According to our forum rules you need to show some effort before receiving help. "I have no idea" is not enough of an effort.
 
Arsene11 said:
Relevant Equations:: F = b + ct^2
"Relevant equations" are standard equations that are likely to have some bearing on the problem, not arbitrary equations specific to the problem.
In this problem you have mass, force, time and velocity mentioned. What standard mechanics equations relate some of those?
 
What is Newton’s 2nd law?

Isn’t it

##F = m \frac{dv}{dt}##

Don’t you have a given expression for ##F##?

Set them equal and solve for ##v##
 
Arsene11 said:
Homework Statement:: A body of mass m = 1 kg moves along a straight line with speed v0, when it is
the action of a force in the same direction of motion. The magnitude of this force is given in terms of
of time t in seconds by the following expression:
F = b + ct^2

where b and which are constants with appropriate relationships for the term-owning unit of
force in Newtons. Find an expression for the velocity v of the body as a function of time t.
Relevant Equations:: F = b + ct^2

I have no idea what to do but the answer is V=V0+ bt +(ct^3)/3
@Arsene11 -- as you have been told, you must show your work before we can be of tutorial help. Please start a new thread and show your best efforts to start working on this problem. Thank you.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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