Asked by Lauren
The initial velocity of a 2.36 kg block sliding down a frictionless inclined plane is found to be 1.46 m/s. Then 1.07 s later, it has a velocity of 4.46 m/s. What is the angle of the plane with respect to the horizontal?
I did this:
a=(4.46-1.46)/1.07=2.80 m/s^2
force along a plane=M*gsinA=M*a
sinA=a/g=2.803/9.8=.286
A=sin^-1(.286)= 16.6 degrees
my homework keeps telling me this:
Use Newton's second law to determine how the forces acting on the block relate to the net force, which is essentially given. If you need the mass of the block in your final calculation, you have made a mistake somewhere.
Please help.
I did this:
a=(4.46-1.46)/1.07=2.80 m/s^2
force along a plane=M*gsinA=M*a
sinA=a/g=2.803/9.8=.286
A=sin^-1(.286)= 16.6 degrees
my homework keeps telling me this:
Use Newton's second law to determine how the forces acting on the block relate to the net force, which is essentially given. If you need the mass of the block in your final calculation, you have made a mistake somewhere.
Please help.
Answers
Answered by
Scott
it appears that the mass of the block is not involved in your final calculation (it cancels out in the second line)
your work seems to be correct
your work seems to be correct
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