What is the acceleration of a crate after it has just started to move?

  • Thread starter Thread starter quanitary
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion focuses on calculating the acceleration of a crate after it begins to move, given the coefficients of static and kinetic friction. The static friction force is determined using the static friction coefficient and the normal force, while the kinetic friction force is calculated with the kinetic friction coefficient. The net force acting on the crate once it starts moving is the difference between the static force and the kinetic friction force. The incorrect initial acceleration calculation of 4.39 m/s² indicates a misunderstanding in applying the forces. Properly accounting for these forces will yield the correct acceleration of the crate.
quanitary
Messages
5
Reaction score
0

Homework Statement



The coefficients of static and kinetic friction between a 445 N crate and the warehouse floor are 0.565 and 0.450, respectively. A worker gradually increases his horizontal push against this crate until it just begins to move and from then on maintains that same maximum push.


Homework Equations


What is the acceleration of a crate after it has just started to move?


The Attempt at a Solution



I tried to multiply the kinectic friction coefficient by the Normal Force to get kinetic friction. Then I tried to divide Kinectic Friction by the mass of the box and I got:

4.39 m/s2 but it was wrong.
 
Physics news on Phys.org
quanitary said:

Homework Statement



The coefficients of static and kinetic friction between a 445 N crate and the warehouse floor are 0.565 and 0.450, respectively. A worker gradually increases his horizontal push against this crate until it just begins to move and from then on maintains that same maximum push.


Homework Equations


What is the acceleration of a crate after it has just started to move?


The Attempt at a Solution



I tried to multiply the kinectic friction coefficient by the Normal Force to get kinetic friction. Then I tried to divide Kinectic Friction by the mass of the box and I got:

4.39 m/s2 but it was wrong.

The force needed to start moving the box is F_static= mu_static * Normal
Once the box starts moving the force of kinetic friction is F_kinetic=mu_kinetic *normal

The net force at that point is F_net=F_static-F-kinetic
a = F_net/mass_crate
 
korican04 said:
The force needed to start moving the box is F_static= mu_static * Normal
Once the box starts moving the force of kinetic friction is F_kinetic=mu_kinetic *normal

The net force at that point is F_net=F_static-F-kinetic
a = F_net/mass_crate

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