Answer:
[tex]W_{fr} = 281.539\,J[/tex]
Explanation:
The kinetic force of friction is:
[tex]f = (0.260)\cdot (31\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot \cos 32^{\textdegree}[/tex]
[tex]f = 67.033\,N[/tex]
Before calculating work, it is require to determine if force exerted on the crate is enough to move it at least. The equation of equilibrium for the crate is:
[tex]\Sigma F = F - f + m\cdot g \cdot \sin \theta = m \cdot a[/tex]
The acceleration experimented by the crate is:
[tex]a = \frac{F-f}{m} + g\cdot \sin \theta[/tex]
[tex]a = \frac{118\,N-67.033\,N}{31\,kg}+(9.807\,\frac{kg}{m^{2}} )\cdot \sin 32^{\textdegree}[/tex]
[tex]a = 6.841\,\frac{m}{s^{2}}[/tex]
This positive result indicates that motion is physically reasonable. Hence, the work done by the force of friction is:
[tex]W_{fr} = (67.033\,N)\cdot (4.20\,m)[/tex]
[tex]W_{fr} = 281.539\,J[/tex]
