Answer:
[tex]a. W_w=235\ J\\b. W_g=-343.54\ J\\c. F_N=463.100\ N\\d. W_t=235\ J[/tex]
Explanation:
Given: that,
Angle of inclination of the surface, [tex]\theta=34^{\circ}[/tex]
mass of the crate, [tex]m=57\ kg[/tex]
Force applied along the surface, [tex]F=230\ N[/tex]
distance the crate moves after the application of force, [tex]s=1.1\ m[/tex]
a) work done = F× s
work done = 230 × 1.1
work done = 253 J
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\ \thetah=1.1\times sin\ 34^{\circ}h=0.615\ m[/tex]
So, the work done by the gravity:
[tex]W_g=57\times 9.8\times (-0.615) \\= - 343.54 J[/tex]
∵direction of force and displacement are opposite.
= - 343.54J
c)
The normal reaction force on the crate by the inclined surface:
[tex]F_N=m.g.cos\ \thetaF_N=57\times 9.8\times cos\ 34F_N=463.100\ N[/tex]
d)
Total work done on crate is with respect to the worker:
[tex]W_t=235\ J[/tex]
