Respuesta :
Answer:
Explanation:
given
m= 17.5kg
F= 75N
d= 5.7m
∪=0.150
θ= 21°
a. W = Fcos θ × d
75cos21° ×5.7
=399.106J
b. normal force is zero. 0 Joules
cos 90°=0
(a) The work done on the block by the 75-N force is [tex]399N[/tex]
(b) The work done on the block by the normal force is 0
(c) The work done on the block by the gravitational force is 0
(d) the increase in internal energy of the block is [tex]252.37J[/tex]
(e) the change in kinetic energy of the block is [tex]252.37J[/tex]
Work Done:
(a) The angle between the force [tex]F=75N[/tex] and horizontal level is [tex]\theta = 21^o[/tex]. Now the work done is given by
[tex]W_F=F.d=Fdcos\theta=75\times5.7\times cos21=399J[/tex]
here d = 5.7 m is the displacement
(b) The normal force is always perpendicular to the horizontal so the angle between the displacement and normal force will be 90° so [tex]cos\theta[/tex] will be zero in this case. therefore work done will be zero.
(c) The gravitational force is also perpendicular to the horizontal so the angle between the displacement and normal force will be 90° so [tex]cos\theta[/tex] will be zero in this case also. therefore work done will be zero.
(d) The frictional force is given by;
[tex]f=\mu N=\mu mg=0.150\times17.5\times9.8=25.725N[/tex]
So the work done is: [tex]W_f=25.725\times5.7\cos(180)=-146.63J[/tex]
The increase in internal energy:
[tex]\DetlaU=W_f+W_F=-146.63+399=252.37J[/tex]
(g) From the work-energy theorem the work done is equal to the change in kinetic energy, therefore:
[tex]\Delta KE=252.37J[/tex]
Learn more about the work-energy theorem:
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