Does V0 affect acceleration under constant Force?

AI Thread Summary
The discussion centers on calculating the acceleration of a block under constant force while considering friction. For a block with a mass of 10 kg and a coefficient of friction of 0.1, the applied force of 5 N results in zero acceleration due to friction being greater than the applied force. In the second scenario with an initial velocity of 2 m/s and the same 5 N force, the user is uncertain about the effect of initial velocity on acceleration. The calculations suggest that the net force determines acceleration, indicating that initial velocity does not influence the acceleration directly. The user seeks clarification on how to compute acceleration in the second case, hinting at a possible acceleration of -0.5 m/s².
VinnieD
Messages
3
Reaction score
0

Homework Statement



A block of mass M = 10 kg is sitting on the table as shwoin Figure above. There is friction between the block and the table, the coefficient of friction is u = 0.1.
The block has initial velocity V, and a constant forceF is applied to the black as showing Figure above.
A) For the case of V = 0 and F = 5 N, find acceleration of the block
B) Constant force is F = 5N, and at time t = 0, the initial velocity is v = 2 m/s, find acceleration of the block.
C) Constant force is F = 15N, and at time t = 0, the initial velocity is v = 2 m/s find acceleration of the block.

It asks to find the dependence of velocity and position as functions of time, but I assume i just integrate to find those.

Homework Equations


ΣF=ma
Ff=μN

The Attempt at a Solution


F0-Ff=10kg(a)
a=(5N-10N)/10kg; a=0

So I think the acceleration is 0 in both cases, but I'm uncertain if the initial velocity would have an effect. Assuming I derived a correctly V0 doesn't affect a, but I'm honestly not sure.
 
Last edited:
Physics news on Phys.org
You didn't state the problem, but in any case acceleration only depends on the net force.
 
VinnieD said:
a=(5N-10N)/10kg; a=0
(5 - 10)/10 does not equal zero.
 
I know in the first case the force of friction is greater than the applied force so its a is 0.
How do I find the acceleration of the second case with the V0 of 2?

Is it just -.5m/s2
 
Please state the complete problem!
 
  • Like
Likes SammyS
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