Efficiency or is calculated by formula:
where
also efficiency has no unit.
Now put in the data and calculate the efficiency.
The efficiency of the machine is 0.8
If you want to get percentage just multiply by 100 and you get 80%
Hope this helps.
r3t40
Going down the first ramp:
• net force parallel to the ramp:
∑ <em>F</em> = <em>W</em> sin(60°) = <em>m</em> <em>a</em>₁
(<em>W</em> for <u>w</u>eight)
• net force perpendicular to the ramp:
∑ <em>F</em> = <em>N</em> + <em>W</em> cos(60°) = 0
(<em>N</em> for <u>n</u>ormal force)
The body has mass 0.1 kg, and with <em>g</em> = 10 m/s², its weight is <em>W</em> = 1 N. So in the first equation, we get
(1 N) sin(60°) = (0.1 kg) <em>a</em>₁ → <em>a</em>₁ ≈ 8.7 m/s²
Let <em>d</em>₁ be the distance the body moves down the ramp, i.e. the distance along the ramp between the starting point and the point <em>O</em>. Then
sin(60°) = <em>h</em> / <em>d</em>₁ → <em>d</em>₁ = 2<em>h</em>/√(3) ≈ 1.15<em>h</em>
Given an initial speed <em>v</em>₁ = 3 m/s, we find the speed <em>v</em>₂ at point <em>O</em> to be
<em>v</em>₂² - (3 m/s)² = 2 (8.7 m/s²) <em>d</em>₁
<em>v</em>₂ ≈ √(9 m²/s² + (17.4 m/s²) (1.15<em>h</em>))
<em>v</em>₂ ≈ √(9 m²/s² + (20 m/s²)<em> h</em>)
Going up the second ramp:
• net parallel force:
∑ <em>F</em> = -<em>Fr</em> - <em>W</em> sin(30°) = <em>m</em> <em>a</em>₂
(<em>Fr</em> for <u>fr</u>iction)
• net perpendicular force:
∑ <em>F</em> = <em>N</em> - <em>W</em> cos(30°) = 0
sin(30°) = cos(60°), and cos(30°) = sin(60°), so the second equation gives <em>N</em> = 0.87 N. Then with <em>µ</em> = 0.1, we have <em>Fr</em> = <em>µ</em> <em>N</em> = 0.087 N. The first equation then gives
-0.087 N - 0.5 N = (0.1 kg) <em>a</em>₂ → <em>a</em>₂ ≈ -5.9 m/s²
We now have
tan(30°) = <em>h</em>/<em>R</em> → <em>h</em> = (2.5 m)/√3 ≈ 1.4 m
(which, by the way, tells us that <em>v</em>₂ ≈ 6.2 m/s)
Then the distance traveled up the ramp is
<em>d</em>₂ = √(<em>h</em>² + <em>R</em> ²) ≈ 2.9 m
Use this to solve for the speed at the top of the ramp:
<em>v</em>₃² - <em>v</em>₂² = 2 (-5.9 m/s²) <em>d</em>₂
<em>v</em>₃ = √((6.2 m/s)² - (11.8 m/s²) (2.9 m)) ≈ 2.0 m/s
At the top of the second ramp:
• net parallel force:
∑ <em>F</em> = -<em>Fsp</em> - <em>W</em> sin(30°) = <em>m a</em>₂
(<em>Fsp</em> for <u>sp</u>ring)
• net perpendicular force:
∑ <em>F</em> = <em>N</em> - <em>W</em> cos(30°) = 0
By Hooke's law, <em>Fsp</em> = <em>kx</em>, so in the first equation we get
-<em>k</em> (0.10 m) - (1 N) cos(60°) = (0.1 kg) (-5.9 m/s²)
→ <em>k</em> ≈ 0.87 N/m
