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Determine the velocity ratio of the machine.

Physics

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Answer

6

η=55.6%\eta = 55.6\%

h=29.4 mh = 29.4 \, \text{m}

Step-by-step solution

4 steps

  1. Determine the velocity ratio of the machine.

    Based on the pulley system shown in the previous question, there are 6 rope segments supporting the movable block.
    Therefore, the velocity ratio (VR) is:

    VR=6\text{VR} = \boxed{6}

  2. Determine the efficiency of the pulley system.

    Given:
    Load, L=500 NL = 500 \, \text{N} (from previous question)
    Effort, E=150 NE = 150 \, \text{N}
    Velocity Ratio, VR = 6

    First, calculate the Mechanical Advantage (MA):

    MA=LoadEffort\text{MA} = \frac{\text{Load}}{\text{Effort}} MA=500 N150 N\text{MA} = \frac{500 \, \text{N}}{150 \, \text{N}} MA=103≈3.333\text{MA} = \frac{10}{3} \approx 3.333

    Next, calculate the efficiency (η\eta):

    η=MAVR×100%\eta = \frac{\text{MA}}{\text{VR}} \times 100\% η=10/36×100%\eta = \frac{10/3}{6} \times 100\% η=1018×100%\eta = \frac{10}{18} \times 100\% η=59×100%\eta = \frac{5}{9} \times 100\% η≈55.55%\eta \approx 55.55\%

    Rounding to three significant figures:

    η=55.6%\eta = \boxed{55.6\%}

  3. Sketch a graph of efficiency against load for the system.

    The provided sketch shows a typical graph of efficiency versus load for a machine. The efficiency generally increases with the load up to a certain point, after which it tends to level off. This is because the constant frictional forces become a smaller proportion of the total work done as the load increases.

  4. Determine how high the block will rise.

    Given:
    Force, F=20 NF = 20 \, \text{N}
    Mass, m=1 kgm = 1 \, \text{kg}
    Time, t=1.2 st = 1.2 \, \text{s}
    Acceleration due to gravity, g=9.8 m/s2g = 9.8 \, \text{m/s}^2

    First, calculate the acceleration aa of the block using Newton's second law:

    F=maF = ma a=Fma = \frac{F}{m} a=20 N1 kga = \frac{20 \, \text{N}}{1 \, \text{kg}} a=20 m/s2a = 20 \, \text{m/s}^2

    Next, calculate the velocity vv of the block after the force acts for 1.2 s1.2 \, \text{s} (assuming it starts from rest):

    v=u+atv = u + at v=0+(20 m/s2)(1.2 s)v = 0 + (20 \, \text{m/s}^2)(1.2 \, \text{s}) v=24 m/sv = 24 \, \text{m/s}

    Finally, use the conservation of energy to find the maximum height hh the block rises. The kinetic energy at the bottom of the swing is converted into potential energy at the maximum height:

    12mv2=mgh\frac{1}{2}mv^2 = mgh

    Divide both sides by mm:

    12v2=gh\frac{1}{2}v^2 = gh

    Solve for hh:

    h=v22gh = \frac{v^2}{2g} h=(24 m/s)22×9.8 m/s2h = \frac{(24 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} h=576 m2/s219.6 m/s2h = \frac{576 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} h≈29.3877 mh \approx 29.3877 \, \text{m}

    Rounding to three significant figures:

    h=29.4 mh = \boxed{29.4 \, \text{m}}
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