Moving Particle is Brought to Rest by a Resisting Force

AI Thread Summary
A particle with mass m is initially traveling at velocity u m/s and is brought to rest by a resisting force defined as km(2u - v). The discussion emphasizes the need to use calculus rather than kinematic equations due to the variable nature of the acceleration. A differential equation, specifically dv/dt = -k(2u - v), is established to model the motion, indicating that the force is retarding. Participants clarify that the goal is to solve this equation to find the time when the particle stops and subsequently calculate the distance traveled during this deceleration. The initial velocity u is constant, while v represents the changing velocity of the particle.
jmlibunao
Messages
16
Reaction score
0

Homework Statement


A particle of mass m kg is traveling in a horizontal straight line with a velocity u m/s. It is brought to rest by means of a resisting force of magnitude km(2u - v), where v is the velocity of the particle at any instant and k is a positive constant.

Find the distance traveled by the particle while v decreases from u top zero


Homework Equations


F = ma
K = (1/2)(m)(v^2)

I think you're also going to need the formula for conservation of energy as well
K1 + E1 = K2 + E2

The Attempt at a Solution


I made this equation F = ma = km(2u - v) and then solved for a as a = k(2u - v)
I tried using the kinematic equation vf = vi + at, where vf = 0 and vi = u and solved for time, t. Then I plugged t into xf = xi + vi(t) + (1/2)(a)(t^2) but I just ended up with an ugly equation filled with variables. I think you have to solve for k but I'm not sure how.

Help would be much appreciated!
 
Physics news on Phys.org
jmlibunao said:

Homework Statement


A particle of mass m kg is traveling in a horizontal straight line with a velocity u m/s. It is brought to rest by means of a resisting force of magnitude km(2u - v), where v is the velocity of the particle at any instant and k is a positive constant.

Find the distance traveled by the particle while v decreases from u top zero


Homework Equations


F = ma
K = (1/2)(m)(v^2)

I think you're also going to need the formula for conservation of energy as well
K1 + E1 = K2 + E2

The Attempt at a Solution


I made this equation F = ma = km(2u - v) and then solved for a as a = k(2u - v)
I tried using the kinematic equation vf = vi + at, where vf = 0 and vi = u and solved for time, t. Then I plugged t into xf = xi + vi(t) + (1/2)(a)(t^2) but I just ended up with an ugly equation filled with variables. I think you have to solve for k but I'm not sure how.

Help would be much appreciated!

Those kinematic equations only apply in constant acceleration scenarios. Here, the force (and hence acceleration) is clearly variable.

You need to use calculus for this. First step: try to express ##v## in terms of ##t## by solving a differential equation. Second step: use that to figure out the time when the particle comes to rest. Final step: use integration to figure out the distance traveled in that time.
 
Thanks for your reply!
I forgot that the kinematic equations are only applicable for constant acceleration.

Curious3141 said:
You need to use calculus for this. First step: try to express v in terms of t by solving a differential equation.

What differential equation are you talking about? dv/dt = k(2u - v) ?? If this is right/wrong, then can you guide me through it? I'm just really having a hard time with this problem.
 
jmlibunao said:
Thanks for your reply!
I forgot that the kinematic equations are only applicable for constant acceleration.



What differential equation are you talking about? dv/dt = k(2u - v) ?? If this is right/wrong, then can you guide me through it? I'm just really having a hard time with this problem.

Don't forget it's a *retarding* force. Hence the equation should be ##\displaystyle \frac{dv}{dt} = -k(2u-v)##. Note the minus sign.

It's a simple ordinary first order differential equation with separable variables - pretty much the most elementary type there is. Have you learned how to solve them? If not, it'll probably take too long (and be ineffective) to instruct you over this forum, so I suggest you do a little reading around the topic.
 
Don't worry about that. I know how to solve differential equations :)

After solving this I'm going to be setting the velocity to 0, right? Which one, u or v, or does it not matter?
 
You say "with a velocity u m/s" and then "v is the velocity of the particle at any instant" so "u" is its initial speed and is constant? In that case v(t) is the variable speed and v should be set to 0.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top