Respuesta :
Answer:
a) 3.6 J
b) 0.9 J
c) 0 J
d) 0 J
e) 2.7 J
Explanation:
Work = force F x diaplament in direction of force d
Given that,
Displacement in direction of push force d = 1.5 N
Push force = 2.4 N
Friction = 0.6 N
a) work done by push force = 2.4 x 1.5 = 3.6 J
b) word done by friction = -0.6 x 15 = -0.9 J (minus sign shows the work is opposite that done by the push force)
c) work done by normal force from table top ( force acting upwards to oppose gravity) = 0 J since there is no vertical displacement
d) work done by gravity (doward force) = 0 J since there is no vertical displacement
e) net work done on book = 3.6 J + (-0.9J) + 0 J + 0 J = 2.7 J
Answer:
a) Work done by the 2.4 N force, W = 3.6 J
b) Work done by the o.6 N frictional force, W = -0.9 J
c) Work done by the normal force on the book, W = 0 J
d) Work done on the book by gravity, W = 0 J
e) Net work done, W = 2.7 J
Explanation:
The distance covered by the book on the horizontal tabletop, s = 1.50 m
Horizontal force applied on the book, f = 2.4 N
The angle between the book and the table top, [tex]\phi = 0^{0}[/tex]
a) Work done by the 2.40 N force on the book,
[tex]W = fscos \phi[/tex]
[tex]W = 2.4 * 1.5cos 0[/tex]
W = 3.6 J
b) Work done by the frictional force, [tex]f_{r} = 0.6 N[/tex]
[tex]W = f_{r} scos \phi[/tex]
Frictional force acts in the opposite direction to the push
[tex]\phi = 180 \\W = 0.6 * 1.5cos 180\\W = -0.9 J[/tex]
c) Work done by the normal force, N on the book
The angle between the normal force and the displacement [tex]\phi = 90^{0}[/tex]
[tex]W = N * scos 90\\W = 0 J[/tex]
d) Work done on the book by gravity
[tex]W = mgs cos \phi[/tex]
Angle between the displacement and gravity, [tex]\phi = 270^{0}[/tex]
[tex]W = mgcos270\\W = 0 J[/tex]
e) Net work done on the book
[tex]W_{net} = 3.6 - 0.9 + 0 + 0\\W_{net} = 2.7 J[/tex]
