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Answer:
d = 57.7 m
Explanation:
For this exercise we will use energy work theorems
W = ΔEm
Let's calculate the work of the rubbing force on the ramp
W = -fr d
The negative sign is because the force of friction opposes the movement
Newton's second law in the axis perpendicular to the ramp gives
N- W cos 21.8 = 0
N = mg cos 21.8
The equation for the force of friction is
fr = μN
fr = μ mg cos 21.8
Let's replace
W = - μ mg d cos 21.8
Let's look for energy in two points
Initial just before entering the ramp
Emo = K = ½ m v²
End the highest point where the body stops
= U = mg y
We use trigonometry to find the height (y)
sin 21.8 = y / d
y = d sin 21.8
= m g d sin 21.8
Let's replace
W = ΔEm = - Em₀
- μ mg d cos 21.8 = mgd sin21.8 - ½ m v²
m g d (sin21.8 + μ cos 21.8) = ½ m v²
d = v² / (2 (sin21.8 + μ cos21.8))
d = 7.10² / (2 (sin21.8 + 0.0704 cos 21.8))
d = 50.41 /0.87345
d = 57.7 m
Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: 
where
is the intensity at distance 2
is the intensity at distance 1
is distance 2
is distance 1
We can infer for our problem that
, 
, and 
. Lets replace those values in our formula to find :
dB
We can conclude that the intensity of the sound when is <span>3 m from the source is 30 dB.</span>
where
We can infer for our problem that
We can conclude that the intensity of the sound when is <span>3 m from the source is 30 dB.</span>