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Sliding block

  • Thread starter jti3066
  • Start date
  • #1
46
0

Homework Statement



A 5.86 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F = 10.00 N at an angle = 35.0° above the horizontal (x-axis). What is the speed of the block 3.70 seconds after it starts moving?

Homework Equations



Fcos() - ma = 0

Fsin() - mg = 0

V_f = V_0 + at (V_0 = 0)

The Attempt at a Solution



Since the first two equations are equal to zero I set them equal to each other and then solved for acceleration. I then used this in the last equation to solve for V_f. Apparently I am doing something wrong because I am not getting the correct answer.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
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What is the value of the acceleration? Will you show your calculations?
 
  • #3
46
0


Fcos() - ma = 0

Fsin() - mg = 0

so setting these two equations together and inputting given data:

10cos(35) - 5.86(a) = 10sin(35) -5.86(9.81)

8.192 - 5.86(a) = 5.736 - 57.487

-5.86(a) = 5.736 - 57.487 - 8.192

-5.86(a) = -59.943

a = 10.22918089 m/s^2

Now to solve for V_f (final velocity)

V_f = V_0 + at

V_f = 0 + 10.23(3.7) = 37.84 m/s

FAIL!!!!!! This is the wrong answer
 
  • #4
rl.bhat
Homework Helper
4,433
7
10cos(35) - 5.86(a) = 10sin(35) -5.86(9.81)

This step is wrong. There is no acceleration in the vertical direction because the action and reaction are equal and opposite. Hence

F*cos(θ) = ma
 
  • #5
46
0
Ok....got it

This always holds true correct, except when the object is inclined/declined

here are the proper calculations:

10cos(35) = 5.86(a)

a = 1.39 m/s^2

V_f = 0 + 1.39(3.7) = 5.17 m/s

correct answer

Thanks a bunch
 

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