b) The work done by friction is -348 J
c) The work done by the puller is 750 J
Explanation:
b)
The force of friction on the crate is given by:
[tex]F_f=\mu N[/tex]
where
[tex]\mu=0.49[/tex] is the coefficient of friction
N is the normal force exerted by the incline on the crate
And the force acts down along the incline.
We need to find N. In order to do that, we have to analyze the forces acting along the direction perpendicular to the incline. We have:
- The normal reaction, N, upward
- The component of the weight perpendicular to the plane, [tex]mg cos \theta[/tex], downward
Since the crate is in equilibrium along the perpendicular direction, we can write:
[tex]N-mg cos \theta = 0[/tex]
where
m = 10.0 kg is the mass of the crate
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\theta=15.0^{\circ}[/tex] is the angle of the incline
Solving for N,
[tex]N=mg cos \theta = (10.0)(9.8)(cos 15.0^{\circ})=94.7 N[/tex]
So now we can find the magnitude of the force of friction:
[tex]F_f=\mu N = (0.49)(94.7)=46.4 N[/tex]
Now we can finally find the work done by the force of friction on the crate:
[tex]W_f = -F_f d = -(46.4)(7.5)=-348 J[/tex]
where the negative sign is due to the fact that the force (down along the incline) is in a direction opposite to the displacement (up along the incline), and where d = 7.5 m is the displacement of the crate.
c)
The work done by the puller on the crate is given by:
[tex]W=Fd[/tex]
where
F is the force applied by the puller
d is the displacement of the crate
In this case, F and d are parallel to each other (they both acts up along the incline), so there is no negative sign
We have:
F = 100.0 N
d = 7.5 m
Substituting, we find:
[tex]W=(100.0)(7.5)=750 J[/tex]
Learn more about work:
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