Answer
$W_{total}=2.484\times 10^3J$
Work Step by Step
We can determine the required work done as follows:
We know that
$f=\mu_k mg\frac{\sqrt{L^2-h^2}}{L}$
We plug in the known values to obtain:
$f=0.2\times 1400\times \frac{\sqrt{(4)^2-(1)^2}}{1}$
This simplifies to:
$f=1084.44N$
Now, the work done by the friction force is given as
$W_{friction}=fh$
We plug in the known values to obtain:
$W_{friction}=1084.44\times 1=1084.44J$
The work done due to the force of gravity is given as
$W_{gravity}=mgh$
We plug in the known values to obtain:
$W_{gravity}=1400\times 1=1400J$
The total work done can be calculated as
$W_{total}=W_{friction}+W_{gravity}$
We plug in the known values to obtain:
$W_{total}=1084.44J+1400J$
$\implies W_{total}=2.484\times 10^3J$
