Question

8. Using a ramp 6 meters long, workers apply an effort force of 1250 N to move a
2000 N (F.) crate onto a platform 2 meters high. What is the efficiency of the
ramp? (Hint: calculate Win & Wout, then efficiency
​

Answer 1

The efficiency of the machine is 0.53 (53%).

Explanation:

The efficiency of a machine is given by:

$\eta = \frac{W_{out}}{W_{in}}$

where

$W_{out}$ is the output work

$W_{in}$ is the work in input

The output work is:

$W_{out}=F_L d_L = (2000)(2)=4000 J$

where

$F_L = 2000 N$ is the load

$d_L = 2 m$ is the distance covered by the load

The input work is:

$W_{in}=F_E d_E = (1250)(6)=7500 J$

where

$F_E = 1250 N$ is the effort force

$d_E = 6 m$ is the distance from the effort to the pivot

Solving for the efficiency,

$\eta=\frac{4000}{7500}=0.53$

So, the efficiency of the machine is 0.53 (53%).

Learn more about work:

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