Motion and Forces

  • Thread starter anastasiaw
  • Start date
  • #1
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A 5.82 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F = 12.80 N at an angle θ = 24.0° above the horizontal as shown. What is the speed of the block 5.90 seconds after it starts moving?

http://img239.imageshack.us/img239/1153/prob49dj7.gif [Broken]

F = ma
Fx = Fcosθ
Fy = Fsinθ
v(t) = v0 + at

Fx = 12.8cos24 = 11.69 N
Fx = m*ax ax = 2.01 m/s^2
vx(t) = 0 + 2.01(5.9) = 11.859 m/s

Fy = 12.8sin24 = 5.21 N
Fy = m*ay ay = 0.895 m/s^2
vy(t) = 0 + (.895)(5.9) = 5.2805 m/s

v(t)^2 = vx(t)^2 + vy(t)^2 = 11.859^2 + 5.2805^2
v(t) = 12.98 m/s
I entered 13.0 m/s into LONCAPA and it told me it was incorrect. What am I doing wrong?

Edit: Also tried a simpler way:

F=ma
12.8=5.82a
a=2.20m/s
v=v0+at
v=0+2.20(5.9)
v=12.98m/s
 
Last edited by a moderator:

Answers and Replies

  • #2
335
0
The second method is satisfactory.

YOU ARE MAKING A SIMPLE MISTAKE:

cos24 degrees=.9135

>>force=12.80 * .9135 = 11.69N

now a= 11.69/5.82 = 2.009 m/s^2

>v= v0+at

>0+2.009*5.90

>v= 11.854 m/s
 

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