Box sliding down plane - find work

In summary, in the first part of the conversation, a 3.00kg box initially at rest slides 1.50m down a frictionless plane inclined at 20 degrees to the horizontal. The work done on the box was 15.08N*m and the final velocity of the box was 3.16m/s. In the second part of the conversation, with a friction coefficient of 0.250, the work done on the box was 3.15N and the final velocity of the box was 2.51m/s.
  • #1
NewJersey
35
0
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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  • #2
In the question above, if the coefficent 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
 
  • #3
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.
 
  • #4
what is mu
and why can you tell all the equation you used.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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"?
 
  • #8
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.
 
  • #9
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 coefficent 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 coefficent 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.
 

Related to Box sliding down plane - find work

1. How do you calculate the work done by a box sliding down a plane?

To calculate the work done by a box sliding down a plane, you first need to find the distance the box slides. Then, multiply the distance by the force of gravity acting on the box (mg) and the cosine of the angle between the plane and the horizontal surface. This will give you the work done in joules (J).

2. What is the formula for work done in this scenario?

The formula for work done is W = Fd cosθ, where W is work (in J), F is the force of gravity (in N), d is the distance (in m), and θ is the angle between the plane and the horizontal surface (in degrees).

3. How does the angle of the plane affect the work done?

The angle of the plane affects the work done by changing the distance the box slides and the angle between the force of gravity and the displacement. The greater the angle, the longer the distance and the smaller the angle between the force and the displacement, resulting in a smaller value for work done.

4. What is the relationship between the work done and the speed of the box?

The work done is directly proportional to the speed of the box. This means that as the speed of the box increases, the work done also increases. This is because the faster the box slides, the greater the distance it covers and the more work is required to overcome the force of gravity.

5. Can the work done by the box be negative?

Yes, the work done by the box can be negative if the box is sliding up the plane instead of down. This is because the force of gravity and the displacement have opposite directions, resulting in a negative value for work done. However, the magnitude of the work done will still be the same as when the box is sliding down the plane.

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