A 200 kg load is raised using a 110 m long lever as shown in the diagram above. The load is 10m from the pivot P. If the efficiency of the the lever is 80%,...

Answer Details

To find the effort E required to lift the load, we first need to understand the concept of mechanical efficiency in levers.
A lever is a simple machine that consists of a rigid beam (lever arm) that pivots around a fixed point called the fulcrum. In this case, the fulcrum is point P.
The mechanical efficiency of a lever is defined as the ratio of the output work done (load lifted) to the input work done (effort applied). Mathematically, it can be expressed as:
Efficiency = (Output Work / Input Work) * 100%
In this problem, the load is the output work and the effort is the input work.
Given: Load = 200 kg Length of lever (distance between fulcrum and load) = 10 m Efficiency = 80% Gravitational acceleration (g) = 10 m/s^2
To calculate the effort, let's first calculate the output work:
Output Work = Load * Distance lifted
The distance lifted is equal to the length of the lever arm, which is 10 m.
Output Work = 200 kg * 10 m = 2000 kg·m
Since 1 kg·m is equivalent to 10 J (1 Joule), we can convert the units:
Output Work = 2000 kg·m * 10 J/kg·m = 20000 J
Now, let's calculate the input work:
Input Work = Effort * Distance moved by the effort
The distance moved by the effort is the length of the lever arm, which is 110 m.
Input Work = Effort * 110 m
Using the formula for mechanical efficiency, we can rewrite it as:
Efficiency = (Output Work / Input Work) * 100%
Solving for the effort:
Effort = (Output Work / (Efficiency/100)) / Distance moved by the effort
Effort = (20000 J / (80/100)) / 110 m
Simplifying the equation:
Effort = (20000 J / 0.8) / 110 m
Effort = 250 J / m
Given that g = 10 m/s^2, we know that 1 N = 1 kg·m/s^2. Therefore, we can convert the units:
Effort = (250 J / m) / (1 kg·m/s^2 / 1 N)
Effort = 250 N
Therefore, the effort E required to lift the load is 250 N.