Answer:
The work done is 8075.2 J.
Explanation:
Given that,
Mass of crate = 160 kg
Distance = 10.3 m
coefficient of friction = 0.50
We need to calculate the normal force
Using balance equation
[tex]F_{y}=ma_{y}[/tex]
[tex]F_{N}-F_{g}=ma_{y}[/tex]
Here, acceleration a = 0
[tex]F_{N}=F_{g}[/tex]
[tex]F_{N}=mg[/tex]
Put the value into the formula
[tex]F_{N}=160\times9.8[/tex]
[tex]F_{N}=1568\ N[/tex]
We need to calculate the frictional force
Using balance equation
[tex]F_{x}=ma_{x}[/tex]
[tex]F_{p}-f_{kx}=0[/tex]
[tex]F_{p}=\mu mg[/tex]
[tex]F_{p}=0.50\times160\times9.8[/tex]
[tex]F_{p}=784\ N[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=F_{p}\ d\cos\theta[/tex]
[tex]W=784\times10.3\cos0[/tex]
[tex]W=8075.2\ J[/tex]
Hence, The work done is 8075.2 J.
