EK Physics 1001 #189

[MaOf2]

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Can someone help me figure out this problem: If theta is 30 degrees m, d is 10 m and x is 10 m, how long after the mass reaches the bottom of the plane does it require to move the distance x?

I can't post the Figure but it is an inclined plane problem. I approached this problem as follows: I know a=g(sin(theta)30degrees), so I took v^2=v(i)^2+2ax and got an answer of v=10m/s. Then I plugged this 10m/s into x=vt+1/2at^2 and started doing the quadratic equation. I think I'm going the wrong way with this. I tried to follow the explanation in the back of the book but couldn't understand it. Can someone please help!!!

Thank you!

 

[mehc012]

Can someone help me figure out this problem: If theta is 30 degrees m, d is 10 m and x is 10 m, how long after the mass reaches the bottom of the plane does it require to move the distance x?

I can't post the Figure but it is an inclined plane problem. I approached this problem as follows: I know a=g(sin(theta)30degrees), so I took v^2=v(i)^2+2ax and got an answer of v=10m/s. Then I plugged this 10m/s into x=vt+1/2at^2 and started doing the quadratic equation. I think I'm going the wrong way with this. I tried to follow the explanation in the back of the book but couldn't understand it. Can someone please help!!!

Thank you!

Friction or no friction? Rolling or sliding?
I would probably use conservation of energy to determine the speed at the bottom of the slope, and then simply use x_f = x₀ + v·t to determine t.

 

[MedSprinter]

Can someone help me figure out this problem: If theta is 30 degrees m, d is 10 m and x is 10 m, how long after the mass reaches the bottom of the plane does it require to move the distance x?

I can't post the Figure but it is an inclined plane problem. I approached this problem as follows: I know a=g(sin(theta)30degrees), so I took v^2=v(i)^2+2ax and got an answer of v=10m/s. Then I plugged this 10m/s into x=vt+1/2at^2 and started doing the quadratic equation. I think I'm going the wrong way with this. I tried to follow the explanation in the back of the book but couldn't understand it. Can someone please help!!!

Thank you!

You are correct that v= 10m/s at the bottom of the inclined plane. Now think about what the acceleration on the flat portion x is. The acceleration on the inclined plane was provided by gravity but once the mass is on the flat surface, the vertical acceleration is irrelevant. Since there is no friction in this problem, the mass does not experience horizontal acceleration. No acceleration= no net force= constant velocity. That means the mass will move across x at 10m/s. Since x is 10m long, it will take 1 second to move across it.

Also, I believe your second equation is incorrect. It should be x= vt-1/2at^2.