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
\(F_{y}=ma_{y}\)
\(F_{N}-F_{g}=ma_{y}\)
Here, acceleration a = 0
\(F_{N}=F_{g}\)
\(F_{N}=mg\)
Put the value into the formula
\(F_{N}=160\times9.8\)
\(F_{N}=1568\ N\)
We need to calculate the frictional force
Using balance equation
\(F_{x}=ma_{x}\)
\(F_{p}-f_{kx}=0\)
\(F_{p}=\mu mg\)
\(F_{p}=0.50\times160\times9.8\)
\(F_{p}=784\ N\)
We need to calculate the work done
Using formula of work done
\(W=F_{p}\ d\cos\theta\)
\(W=784\times10.3\cos0\)
\(W=8075.2\ J\)
Hence, The work done is 8075.2 J.
