Kinetic Friction of shoved box of books

AI Thread Summary
A box of books weighing 300 N is pushed with a 400 N force at a 35.2-degree angle, and the coefficient of kinetic friction is 0.570. The equations of motion were set up to find the acceleration, resulting in an acceleration of -3.83 m/s². Using this acceleration, the time to move 4.00 m from rest was calculated to be 1.44 seconds. A suggestion was made to draw a free body diagram to clarify the forces acting on the box. The discussion emphasizes the importance of accurately applying physics equations to solve the problem.
BBallman_08
Messages
27
Reaction score
0

Homework Statement



A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is .570, how long does it take the box to move 4.00 M, starting from rest.


Homework Equations



E Fx = 400 cos (35.2) - mK Fn = 300 (a)

E Fy = Fn - Fmg - 400 sin (35.2)



The Attempt at a Solution



400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

A= -3.83 m/s2

4m = (1/2) (-3.83) (t^2)

t= 1.44s







This is what I have so far... I don't know if its correct?? Can someone help me out and check it? Thanks!
 
Physics news on Phys.org
BBallman_08 said:

Homework Statement



A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is .570, how long does it take the box to move 4.00 M, starting from rest.


Homework Equations



E Fx = 400 cos (35.2) - mK Fn = 300 (a)

E Fy = Fn - Fmg - 400 sin (35.2)



The Attempt at a Solution



400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

A= -3.83 m/s2

4m = (1/2) (-3.83) (t^2)

t= 1.44s







This is what I have so far... I don't know if its correct?? Can someone help me out and check it? Thanks!


400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

This equation looks funny to me.. can you elaborate on how you got this?

[Edit]

I forgot to mention this, but on these type of problems, it is essential that you draw a free body diagram. Scan and attach one if you drew it.
 
Last edited by a moderator:
I substituted arranged the E Fy to solve for FN and then substituted that in for the FN in my Efx equation that I have listed up top.

Thanks for the help! I don't have a scanner to put my sketch on but I have the applied force coming in from the left side from above the block at an angle of 30 degrees below horizontal, the normal force going up from the box and the Fgrav headed down. The last force I ID'd was the F kf going to the left from the box.
 
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