The floor of a lorry is 2.0m high. A plank 5m long is used as an inclined plane to raise some load up the lorry. If the efficiency of this machine is 50% what is the minimum effort (applied parallel to the plane) required to raise 200N load up the plane?
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work done to raise 200 N up 2 meters = 400 Joules
work needed at 50% efficiency = 800 Joules
distance pushed = 5 meters
so 5 F = 800
F = (800/5) Newtons
5/2=2.5 efficiency=ma/vr×100% 50%=ma/2.5×100ma=50×2.5divided by 100=1.25 therefore 1.25=load/effort=200/effort effort=200/1.25=160Nanswer
the flour of a long is 2.0 high. A plank 5m is used as an inclined to plane to raise some load up the lorry it the efficiency of this machine is 50% what is the maximum effort applied parallel to plane required 10 raised 200n load up the plane
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To find the minimum effort required to raise a 200N load up the inclined plane, we need to consider the efficiency of the machine.
Efficiency is defined as the ratio of output work to input work. In this case, the output work is the work done to raise the load up the lorry, and the input work is the work done by the effort applied parallel to the plane.
Efficiency = (Output Work / Input Work) * 100
Given that the efficiency is 50% (or 0.5 as a decimal), we can use this information to find the minimum effort.
Now, let's calculate the work done while raising the load using the inclined plane.
Work = Force * Distance
Here, the force exerted by the effort is the minimum effort required to raise the load and the distance is the length of the inclined plane.
The length of the inclined plane is given as 5m.
To calculate the output work, we need to find the vertical distance the load is raised.
The vertical distance raised is the height of the lorry's floor, which is given as 2.0m.
Now, we can calculate the output work done:
Output Work = Force * Distance
= 200N * 2.0m
= 400 Joules
Since the efficiency is given as 50%, the output work will be equal to 50% of the input work.
Now, let's calculate the minimum effort required:
Input Work = Output Work / (Efficiency * 100)
= 400 Joules / (0.5 * 100)
= 400 Joules / 50
= 8 Joules
Therefore, the minimum effort required to raise the 200N load up the inclined plane is 8 Joules, applied parallel to the plane.