Respuesta :
Answer:
a. [tex]W_w=313.5\ J[/tex]
b. [tex]W_g=-155.312\ J[/tex]
c. [tex]F_N=222.045\ N[/tex]
d. [tex]W_t=313.5\ J[/tex]
Explanation:
Given:
- angle of inclination of the surface,[tex]\theta=25^{\circ}[/tex]
- mass of the crate, [tex]m=25\ kg[/tex]
- Force applied along the surface, [tex]F=209\ N[/tex]
- distance the crate slides after the application of force, [tex]s=1.5\ m[/tex]
a.
Work done by the worker who applied the force:
[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.
[tex]W_w=209\times 1.5[/tex]
[tex]W_w=313.5\ J[/tex]
b.
Work done by the gravitational force:
[tex]W_g=m.g\times h[/tex]
where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:
[tex]h=s\times sin\ \theta[/tex]
[tex]h=1.5\times sin\ 25^{\circ}[/tex]
[tex]h=0.634\ m[/tex]
So, the work done by the gravity:
[tex]W_g=25\times 9.8\times (-0.634)[/tex] ∵direction of force and displacement are opposite.
[tex]W_g=-155.312\ J[/tex]
c.
The normal reaction force on the crate by the inclined surface:
[tex]F_N=m.g.cos\ \theta[/tex]
[tex]F_N=25\times 9.8\times cos\ 25[/tex]
[tex]F_N=222.045\ N[/tex]
d.
Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]
