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A motor consumes 36000 joules of electrical energy while performing 7000 joules of useful mechanical work. Calculate: a) The efficiency of the motor b) The energy lost

Physics

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Answers

5 parts
a

4905 N\text{4905 N}

b
2.725
c
4
d

68.125%68.125\%

e

18000 J18000 \text{ J}

Step-by-step solution

10 steps

Here's the solution for the inclined plane problem:

Part a· 2 steps

Calculate the load due to the weight of the trolley.

  1. Identify the given mass and the acceleration due to gravity.

    Mass m=500 kgm = 500 \text{ kg}
    Acceleration due to gravity g=9.81 m/s2g = 9.81 \text{ m/s}^2

  2. Calculate the load (weight) using the formula W=mgW = mg.

    W=500 kg×9.81 m/s2W = 500 \text{ kg} \times 9.81 \text{ m/s}^2 W=4905 NW = 4905 \text{ N}

    The load due to the weight of the trolley is 4905 N\boxed{\text{4905 N}}.

Part b2.725

Mechanical advantage (M.A.).

  1. Identify the load and the effort.

    Load W=4905 NW = 4905 \text{ N} (from part a)
    Effort E=1800 NE = 1800 \text{ N}

  2. Calculate the mechanical advantage using the formula M.A.=LoadEffortM.A. = \frac{\text{Load}}{\text{Effort}}.

    M.A.=4905 N1800 NM.A. = \frac{4905 \text{ N}}{1800 \text{ N}} M.A.=2.725M.A. = 2.725

    The mechanical advantage is 2.725\boxed{2.725}.

Part c4

Velocity ratio (V.R.).

  1. Identify the effort distance and the load distance.

    Effort distance (length of inclined plane) L=10 mL = 10 \text{ m}
    Load distance (vertical height) h=2.5 mh = 2.5 \text{ m}

  2. Calculate the velocity ratio using the formula V.R.=Effort distanceLoad distanceV.R. = \frac{\text{Effort distance}}{\text{Load distance}}.

    V.R.=10 m2.5 mV.R. = \frac{10 \text{ m}}{2.5 \text{ m}} V.R.=4V.R. = 4

    The velocity ratio is 4\boxed{4}.

Part d

68.125%68.125\%

The efficiency of the inclined plane.

  1. Identify the mechanical advantage and the velocity ratio.

    Mechanical Advantage M.A.=2.725M.A. = 2.725 (from part b)
    Velocity Ratio V.R.=4V.R. = 4 (from part c)

  2. Calculate the efficiency using the formula Efficiency=M.A.V.R.×100%\text{Efficiency} = \frac{M.A.}{V.R.} \times 100\%.

    Efficiency=2.7254×100%\text{Efficiency} = \frac{2.725}{4} \times 100\% Efficiency=0.68125×100%\text{Efficiency} = 0.68125 \times 100\% Efficiency=68.125%\text{Efficiency} = 68.125\%

    The efficiency of the inclined plane is 68.125%\boxed{68.125\%}.

Part e

18000 J18000 \text{ J}

The input work.

  1. Identify the effort and the effort distance.

    Effort E=1800 NE = 1800 \text{ N}
    Effort distance (length of inclined plane) L=10 mL = 10 \text{ m}

  2. Calculate the input work using the formula Input work=Effort×Effort distance\text{Input work} = \text{Effort} \times \text{Effort distance}.

    Input work=1800 N×10 m\text{Input work} = 1800 \text{ N} \times 10 \text{ m} Input work=18000 J\text{Input work} = 18000 \text{ J}

    The input work is 18000 J\boxed{18000 \text{ J}}.

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