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Expert: James J. Kovalcin - 10/31/2009
Question
A 2.4 kg block slides down a 25° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity is 0.55 m/s. The incline is 1.6 m long.
(a) What is the acceleration of the block?
Answer
This is a straightforward kinematics problem with 6 unknowns and 2 equations. Any time you can identify 4 of the 6 variables you can always identify the other 2. In this case the 6 variables are: do=0m [making the starting point the origin], df=1.6m [given - the final displacement of the block when it reaches the bottom of the incline, Vo=0m/s [given - the block starts from rest, Vf=0.55m/s [given], a=? [unknown] and t=? [unknown].
The 2 available kinematics problems are:
df=1/2*a*t^2+Vo*t+do and Vf=a*t+Vo
Which in this case become:
1.6=1/2*a*t^2+0+0 and 0.55=a*t+0
Rewriting the equations:
3.2=[a*t]*t and 0.55=a*t
Replacing a*t in the displacement equation:
3.2=[0.55]*t
And solve for t=3.2/0.55=5.8s
Finally solve for the acceleration using the velocity equation:
0.55=a*t becomes a=0.55/5.8=0.095m/s^2
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