How much work is done on the box by person pulling?

  • Thread starter Thread starter Bones
  • Start date Start date
  • Tags Tags
    Box Work
AI Thread Summary
The discussion focuses on calculating the work done on a box being pulled at an angle, considering various forces involved. The person pulls with a force of 29 N at a 30-degree angle, and the box has a mass of 4.00 kg with friction coefficients of static and kinetic friction. The user successfully calculated the work done by the person pulling as 82.9 J but is struggling with the work done by friction. They are attempting to apply the correct formula for frictional work but are uncertain about their calculations. The conversation emphasizes the need to accurately account for the forces acting on the box to determine the net work done.
Bones
Messages
108
Reaction score
0

Homework Statement



In the figure below,

http://www.webassign.net/userimages/jshemwell@lincolnpark.il/Net%20Force/Pulling_at_angle.gif

the mass is 4.00 kg, and = 30 degrees. The surface does have friction, with s = .45 and k = .39.

If a person takes hold of the rope and, pulling at this angle with a force of T = 29 N, drags the box 3.3 meters...

a. How much work is done on the box by person pulling?

b. How much work is done on the box by the normal force?

c. How much work is done on the box by the force of friction?

d. What is the net work done on the box?

e. If the block starts from rest, what will be its velocity just as the person finishes pulling?

Homework Equations



The Attempt at a Solution



a) 29N*3.3m*cos30 = 82.9 J
b) 0
c) I took 4kg*9.8m/s^2*0.39*3.3m*cos180 = -50.45 which is not correct. What am I doing wrong??
 
Last edited:
Physics news on Phys.org
I can figure out the rest, I just need to know what I am doing wrong in part c. I am stuck on part c.
 
Bones said:
c. How much work is done on the box by the force of friction?

The force of friction over the distance.

W = F * d = u*(m*g - T*sinθ ) * d
 
Thanks again!
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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