Friction problem from Halliday and Resnick

AI Thread Summary
The discussion revolves around a physics problem involving a 1000kg boat decelerating due to friction when its engine is turned off. The frictional force is proportional to the boat's speed, leading to a non-constant acceleration scenario. The user seeks help in formulating the problem into a differential equation to determine the time required for the boat to slow from 90 km/h to 45 km/h. They express confusion over integrating the equation and understanding the relationship between velocity and time. The conversation highlights the need for a better grasp of differential equations to solve such problems effectively.
Farnak
Messages
4
Reaction score
0
Hi, new member here ^_^

"A 1000kg boat is traveling at 90km/h when its engine is shut off. The magnitude of the frictional force f_k between boat and water is proportional to the speed v of the boat: f_k = 70v, where v is in meters per second and f_k is in Newtons. Find the time required for the boat to slow to 45 km/h."

From what I'm getting, the only force on the boat is the force of kinetic friction and as time passes the velocity decreases, so the force of kinetic friction's magnitude will decrease as well. So this will decrease the magnitude of the acceleration and cause velocity to decrease more slowly as time passes by ...

I'm having trouble putting all that together not being used to dealing with non-constant acceleration problems so could someone please help me organize my thoughts for this problem? Thanks!
 
Physics news on Phys.org
This can be summarized neatly in the differential equation,

F = -kv = ma = m\frac{dv}{dt}

where k is the constant of proportionality in the problem. Along with the initial condition v(0) = 90km/h, this equation can be solved to give you v(t). Would you know how to do that? Does this make sense to you?
 
Are you supposed to solve the equation through integration? If so, I think I'm messing up:

-kv = m (dv/dt)
Integrating both sides
(-k/2)v^2 + C = mv

But an equation like this gives me a constant value for velocity =( ... where did I mess up?
 
Farnak said:
-kv = m (dv/dt)
Integrating both sides
(-k/2)v^2 + C = mv

-kv=mdv/dt
therefore b*dv/v=dt where b= -m/k
therefore b*loge(vf/vi)=t
where vf and vi are the final n initial velocities...
 
Ahhh ok, so I guess I need to learn differential equations if I want to read more, sorry for the calculus stupidity, thanks!
 
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