Asked by Emily
The initial velocity of a 2.75 kg block sliding down a frictionless inclined plane is found to be 1.25 m/s. Then 1.05 s later, it has a velocity of 5.31 m/s. What is the angle of the plane with respect to the horizontal?
I keep trying to use sine and cosine but I can't figure it out. Help! Thanks
I keep trying to use sine and cosine but I can't figure it out. Help! Thanks
Answers
Answered by
Scott
the acceleration is
... (5.31 - 1.25) / 1.05
the sine of the angle of inclination is equal to the actual acceleration
divided by gravitational acceleration
sin(Θ) = [(5.31 - 1.25) / 1.05] / g
... (5.31 - 1.25) / 1.05
the sine of the angle of inclination is equal to the actual acceleration
divided by gravitational acceleration
sin(Θ) = [(5.31 - 1.25) / 1.05] / g
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