Box sliding down plane - find work

[NewJersey]

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

 

[NewJersey]

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

 

[Shooting Star]

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.

 

[NewJersey]

what is mu
and why can you tell all the equation you used.

 

[Shooting Star]

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.

 

[NewJersey]

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?

 

[HallsofIvy]

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"?

 

[aq1q]

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.

 

[Shooting Star]

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.

 

[NewJersey]

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

 

[NewJersey]

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

 

[Shooting Star]

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.

 

[Shooting Star]

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.

 

[NewJersey]

I was following a homework example, I must have wrote down N as a mistake.

 

[Shooting Star]

Well, correct it. The force is 3.15. Then proceed as in the first part.