Finding the velocity of an inclined mass

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Homework Help Overview

The problem involves determining the velocity of a mass sliding down an incline at an angle of 60°. The mass starts from rest, and the discussion includes the effects of friction, represented by a kinetic friction coefficient of 0.1.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of energy conservation principles and Newton's second law to find the acceleration and subsequently the velocity. Some question the initial approach and suggest using kinematics after determining acceleration.

Discussion Status

There are multiple interpretations of the problem, with participants discussing different methods to approach the solution. Some guidance is offered regarding the use of Newton's second law, and attempts to clarify the application of forces are noted.

Contextual Notes

Participants are navigating the complexities of friction and the relationship between forces acting on the mass. There is an acknowledgment of potential misapplication of laws in the initial attempts.

Sall1230
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Homework Statement


As shown in the figure, ( a mass sliding down an incline of 60°) a car of 1000 kg mass start from the rest, it's velocity after 5s is? ( the kinetic fraction coefficient 0.1)

Homework Equations


E1=E2
K1+U1=K2+U2
E1+Wf=E2

The Attempt at a Solution


-Fk= mg cos 60 (0.1) = -490
K1+U1 + Wf = K2 + U2
We should cancel K1 and U2 because they equal zero
Then
Mgh -Fk d = 1/2 mv^2
Cancelling both m from each side
gh -Fk (vt) = 1/2 v^2
9.8*h+490(5v) = 1/2 v^2

Then I don't know what to do
 
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Since you want the speed at a given time, why not find the acceleration and then use kinematics. Start with Newton's 2nd law.
 
Doc Al said:
Since you want the speed at a given time, why not find the acceleration and then use kinematics. Start with Newton's 2nd law.

I was doing the wrong law for this equation.

mg sin 60 - Fk = ma
1000*9.8*sin60 - ( 0.1 * 1000*9.8*cos60)= 1000a

8487.07-490=1000a
7997.04/1000
= 40m/s
 
Sall1230 said:
I was doing the wrong law for this equation.

mg sin 60 - Fk = ma
1000*9.8*sin60 - ( 0.1 * 1000*9.8*cos60)= 1000a

8487.07-490=1000a
7997.04/1000
= 40m/s
Good.
 

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