Box sliding down plane - find work

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Homework Help Overview

The discussion revolves around a physics problem involving a box sliding down an inclined plane. The box has a mass of 3.00 kg and slides 1.50 m down a plane inclined at 20 degrees to the horizontal. The problem is divided into two parts: the first part assumes a frictionless scenario, while the second part introduces a coefficient of friction of 0.250.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of work done on the box and its final velocity under different conditions (frictionless vs. frictional). There are attempts to clarify the forces acting on the box, including gravitational components and frictional forces. Questions arise regarding the correct interpretation of the problem and the definitions of variables such as the coefficient of friction.

Discussion Status

The discussion is active with participants providing various approaches to the problem. Some participants have offered guidance on the forces involved, while others are questioning assumptions and clarifying the setup of the problem. There is no explicit consensus on the calculations, but multiple interpretations and methods are being explored.

Contextual Notes

There is some confusion regarding the presence of friction in the first part of the problem, with participants debating whether the plane is frictionless or not. Additionally, there are discrepancies in the values used for acceleration and force calculations, leading to further questions about the setup and assumptions made in the problem.

NewJersey
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A 3.00kg box, initially at rest, slides 1.50m down a frictional plane inclined at 20degrees to the horizontal?



a) The work done on the box was ?
B) the final velocity of the box?




a= 3kg*9.8m/s *sin(20)=10.05N
10.05N *1.5m= 15.08N*m

b= square root of 2* 10/3 * 1.5 = 3.16m/s
 
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In the question above, if the coefficient of friction is 0.250
a) The work done on the box was?
b)the final velocity of the box was ?

Do I set it up 3*g*cos(20)= frictional
 
You have written "a= 3kg*9.8m/s *sin(20)=10.05N", which is not correct. This is not accn, but the component of mg along the plane.

Total force along the plane should be mg*sin(20)-(mu)mg*cos(20). Now you can find W.
 
what is mu
and why can you tell all the equation you used.
 
mu, pronounced mew, is the greek letter we use to denote the co-eff of friction. It's not m into u. We'll use 'k' for it hence.

Frictional force F = kN, where N is the normal reaction at the point of contact.

Normal reaction is mgcos(20 deg) in this case. Total force P acting on the body along the plane is component of weight along the palne minus frictional force. So,
P= mg*sin 20 – kmg*sin 20.

W = F*d.

I am sure you know the formula for final velo and accn. Find the accn and apply it.
 
Ok in the first question , you said used the equation
mg*sin(20)-(mu)mg*cos(20). And mu=frictional force. However in the first equation we are to asumed there is no frictional force because it is no stated. So this equation is used for the second question where frictional force is given?
 
NewJersey said:
Ok in the first question , you said used the equation
mg*sin(20)-(mu)mg*cos(20). And mu=frictional force. However in the first equation we are to asumed there is no frictional force because it is no stated. So this equation is used for the second question where frictional force is given?
?? Why do you say there is no frictional force? What you wrote was, "A 3.00kg box, initially at rest, slides 1.50m down a frictional plane inclined at 20degrees to the horizontal?" Did you mean "frictionless"?
 
for the second part, you should look carefully at what the acceleration should be.. is it really 9.8 or 10(as u wrote it) or something else, maybe a component of gravity.
 
NewJersey said:
A 3.00kg box, initially at rest, slides 1.50m down a frictional plane inclined at 20degrees to the horizontal?

a) The work done on the box was ?
B) the final velocity of the box?

You have not given the co-eff of friction. Is it a frictionless plane? In that case, k=0. The component of the force along the plane you know. Divide my m to get the accn a. You also know the initial velo.
 
  • #10
Okay, I am sorry for the confusion. This is a two part question the first part is frictionless, and it states..

A 3.00kg box, initially at rest, slides 1.50m down a frictionless plane inclined at 20degrees to the horizontal?

a) The work done on the box was ?
B) the final velocity of the box?

and I set it up like this

part a=
3kg*9.8m/s *sin(20)=10.05N
10.05N *1.5m= 15.08N*m

b= square root of 2* 10/3 * 1.5 = 3.16m/s
 
  • #11
The second part has friction and the question is

In the question above, if the coefficient of friction is 0.250
a) The work done on the box was?
b)the final velocity of the box was ?

And I set it up like

part a =
3kg*9.87*sin(20) = 10.05
3kg*9.87*cos(20)= 27.62

Un= (.250)*(27.62)= 6.9
10.05-6.9-3.15

N= 3.15/(.250)= 12.6N


part b

v^2= vo^2+ 2aX

v= square root of 12.6/3 *1.5= 2.51


is this right
 
  • #12
NewJersey said:
Okay, I am sorry for the confusion. This is a two part question the first part is frictionless, and it states..

A 3.00kg box, initially at rest, slides 1.50m down a frictionless plane inclined at 20degrees to the horizontal?

a) The work done on the box was ?
B) the final velocity of the box?

and I set it up like this

part a=
3kg*9.8m/s *sin(20)=10.05N
10.05N *1.5m= 15.08N*m

b= square root of 2* 10/3 * 1.5 = 3.16m/s


This one seems all right. Why 10 instead of 10.05? But let it go.
 
  • #13
NewJersey said:
The second part has friction and the question is

In the question above, if the coefficient of friction is 0.250
a) The work done on the box was?
b)the final velocity of the box was ?

And I set it up like

part a =
3kg*9.87*sin(20) = 10.05
3kg*9.87*cos(20)= 27.62

Un= (.250)*(27.62)= 6.9
10.05-6.9-3.15

N= 3.15/(.250)= 12.6N

What is N? You got the force as 3.15.
 
  • #14
I was following a homework example, I must have wrote down N as a mistake.
 
  • #15
Well, correct it. The force is 3.15. Then proceed as in the first part.
 

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