What distance does the block slide?

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The discussion centers on a physics problem involving the sliding distance of a block on an incline. The user correctly calculated the velocity at 20 m/s but miscalculated the distance, obtaining 0.76 m instead of the expected 32 m. They attempted to apply energy conservation principles and kinematics equations but struggled with the correct formulation, particularly regarding the sine and friction components. The user is seeking clarification on the proper equations to use and where their calculations went wrong. The conversation highlights the complexities of applying physics equations in problem-solving scenarios.
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Homework Statement


upload_2016-5-24_16-24-33.png


Homework Equations


m*v^2 / 2 and k*x^2 / 2 and 0,5*m*v^2

The Attempt at a Solution


[/B]
in task a I got correct answer there v = 20 m/s
in task b I got wrong answer I got the distance = 0,76 m but I should be 32 m

here is my answer :
0,5*m*(sin(30)*v) - mu * m*g*cos(30) = m*g*h
h=0,38 m
sin (30) = h/L there L is the distance of the block when it goes up
L= h/sin(30) = 0,76

What I do wrong in task b ?

Thanks

 
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JiJiasd said:
0,5*m*(sin(30)*v)
sin(30)v should be squared
 
I seem to get the answer on the second one to be 35.4012435624
 
JiJiasd said:
mu * m*g*cos(30)
Shouldnt this be mu*(m*g*cos(30))*h/cos(30)=mu*m*g*h?
I think so...
 
sidt36 said:
I seem to get the answer on the second one to be 35.4012435624
What equation you did use ? thanks
 
Replusz said:
sin(30)v should be squared
I tried with 0,5*m*(sin(30)*v)^2 - mu * m*g*cos(30) = m*g*h
but it does not give me the correct answer
thanks
 
Replusz said:
Shouldnt this be mu*(m*g*cos(30))*h/cos(30)=mu*m*g*h?
I think so...
Now I tried 0,5*m*(sin(30)*v)^2 - mu * m*g*cos(30) *h/sin(30) = m*g*h
but not working
 
I used a kinematics equation
 

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