Answer:
The work done by the 83.1 N force is 4687.5 J.
The magnitude of the work done by the force of friction is 1187.5 J.
Explanation:
Given:
Mass of the block, [tex]m=15.7[/tex] kg
Force acting on it, [tex]F=83.1[/tex] N
Angle of application of force, [tex]\theta = 25.7[/tex]°
Displacement of the block, [tex]d=62.6[/tex] m
Coefficient of friction, [tex]\mu =0.161[/tex]
Acceleration due to gravity, [tex]g=9.8[/tex] m/s²
Work done by a force is given as:
[tex]Work,W=F\times d\times \cos\theta[/tex]
So, work done by the constant force is given as:
[tex]W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}[/tex]
Now, in order to find work done by friction, we need to evaluate friction.
For the vertical direction, the net force is zero as there is no vertical motion.
Therefore,
[tex]N + F\sin\theta = mg\\ N = mg-F\sin\theta[/tex]
Frictional force is given as:
[tex]f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}[/tex]
Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.
Therefore, work done by friction is:
[tex]W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}[/tex]
Negative sign indicates that frictional force is acting opposite to motion.
So, the magnitude of the work done by the force of friction is 1187.5 J.
