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Block pulled at an angle with friction

  • Thread starter sentinel7e
  • Start date
  • #1

Homework Statement



A 4.62 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 12.0 N at an angle θ = 33.0° above the horizontal.

The coefficient of kinetic friction between the block and the floor is 0.100. What is the speed of the block 3.10 s after it starts moving?

Homework Equations



Fn = Uk * (m*g - F*sin[tex]\theta[/tex])

F = ma

The Attempt at a Solution



I solved for Fn and got Fn = 3.87. I assume I need to solve for acceleration and then get velocity but I am not sure.
 

Answers and Replies

  • #2
PhanthomJay
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Homework Statement



A 4.62 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 12.0 N at an angle θ = 33.0° above the horizontal.

The coefficient of kinetic friction between the block and the floor is 0.100. What is the speed of the block 3.10 s after it starts moving?

Homework Equations



Fn = Uk * (m*g - F*sin[tex]\theta[/tex])
You mean that's the friction force
F = ma

The Attempt at a Solution



I solved for Fn and got Fn = 3.87. I assume I need to solve for acceleration and then get velocity but I am not sure.
You correctly solved for the friction force, so yes, continue to solve for the acc. in the horiz direction.
 
  • #3
I'm still stuck, do I use F = ma to find accel and my F = 3.87? I tried it that way but got the wrong answer.
 
  • #4
PhanthomJay
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I'm still stuck, do I use F = ma to find accel and my F = 3.87? I tried it that way but got the wrong answer.
No, its F_net =ma. You have the friction force, 3.87 N, acting horizontally in one direction, and the horiz component of the cord tension force, F, pulling horizontally in the other direction. Calculate the horiz comp of the 12 N force, then solve for the net force and continue.
 
  • #5
Ok, I finally understood what you meant about if being friction force and not Fn. After applying the forces in both directions I was able to get the accel and ultimately the velocity of 4.15 m/s. Thanks for the help!
 

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