A block is released from rest on an inclined plane and moves 3.2 m during the next 3.9 s. The acceleration of gravity is 9.8 m/s2 .
What is the magnitude of the acceleration of the block? Answer in units of m/s2.
The block is 12kg and the incline is 36 degrees.
5 answers
I tried this problem with a=2(l)/t^2, a=2(3.2)/3.95^2 = 0.41019 and the answer is not correct and I am not sure what is wrong with it or how to do it.
Can someone please walk me through what equation I need to use to do this?
find the acceleration down the block:
vf^2=1/2 acceleration*distance
vf=2*avgvelociyt=2*3.2/3.9
putting this together, solve for acceleaetion
a= 2*(2*3.2/3.9)^2 /3.9
vf^2=1/2 acceleration*distance
vf=2*avgvelociyt=2*3.2/3.9
putting this together, solve for acceleaetion
a= 2*(2*3.2/3.9)^2 /3.9
h = 3.2*sin36 = 1.88 m. Ht. of incline.
V^2 = Vo^2 + 2g*h = 0 + 19.6*1.88 = 36.9
V = 6.1 m/s = Final velocity.
V^2 = Vo^2 + 2a*d.
6.1^2 = 0 + 2*a*3.2,
a = 5.8 m/s^2.
Another Method:
M*g = 12*9.8 = 117.6 N. = Wt. of block.
Fp = 117.6*sin36 = 69.1 N. = Force parallel to the incline.
Fp = M*a.
69.1 = 12*a,
a = 5.8 m/s^2.
V^2 = Vo^2 + 2g*h = 0 + 19.6*1.88 = 36.9
V = 6.1 m/s = Final velocity.
V^2 = Vo^2 + 2a*d.
6.1^2 = 0 + 2*a*3.2,
a = 5.8 m/s^2.
Another Method:
M*g = 12*9.8 = 117.6 N. = Wt. of block.
Fp = 117.6*sin36 = 69.1 N. = Force parallel to the incline.
Fp = M*a.
69.1 = 12*a,
a = 5.8 m/s^2.
post it.
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