- #1
bearhug
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A 41 kg box initially at rest is pushed 4 m along a rough, horizontal floor with a constant applied force of 121 N. If the coefficient of kinetic friction between the box and the floor is 0.29, find
(a) the work done by the applied force W= FΔx
Ans: 4.84e2 J
b) the increase in internal energy of the box + floor system due to friction ΔEint= fkd fk= ukn= 116.5N
ΔEint= 4.66e2 J
c)the work done by the normal force on the block due to the floor
=0.00 J
d) the work done by the gravitation force on the block due to the Earth =0.00 J
e)the change in kinetic energy of the box
ΔK = -fkd + ∑W This is where I'm stuck. Is ∑W= to 4.84e2J because I thought that the applied force was included in the change in kinetic energy but this according to my homework is wrong. Can someone explain what I'm missing here?
(a) the work done by the applied force W= FΔx
Ans: 4.84e2 J
b) the increase in internal energy of the box + floor system due to friction ΔEint= fkd fk= ukn= 116.5N
ΔEint= 4.66e2 J
c)the work done by the normal force on the block due to the floor
=0.00 J
d) the work done by the gravitation force on the block due to the Earth =0.00 J
e)the change in kinetic energy of the box
ΔK = -fkd + ∑W This is where I'm stuck. Is ∑W= to 4.84e2J because I thought that the applied force was included in the change in kinetic energy but this according to my homework is wrong. Can someone explain what I'm missing here?