Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
At the end of 4.0 s the velocity of the rock is 39.24 m/s and its displacement from the point where it was dropped is 78.48 m.
The rock dropped from the guard rail is a freely falling body and is accelerated downwards under acceleration due to gravity.
To calculate the velocity after t=4.0 s, use the equation of motion,
Here, <em>u </em>is the rock's initial velocity, which is zero. It falls under acceleration due to gravity, hence a =g=9.81 m/s².
Therefore,
If it falls through a distance <em>s, </em>then, use the equation,
Therefore,
Thus, after 4.0 s, the rock has a downward velocity of 39.24 m/s and it would have traveled a distance of 78.48m.