Respuesta :
Answer:
(a) W = 645 J
(b) E= 461.9 J
(c) The work done by the normal force = 0 J as no displacement is taking place in that direction
(d)The work done by the gravitational force 0 J, as there is zero displacement in that direction too.
(e) ΔKE= 183 J
(f) v = 3.16 m/s
Explanation:
A 36.5 kg box initially at rest is pushed 4.3 m along a rough, horizontal floor with a constant applied horizontal force of 150 N. If the coefficient of friction between box and floor is 0.300, find the following.
(a) the work done by the applied force
W=F*d
W=150*4.3
W = 645 J
(b)
The increase in internal energy in the box-floor system due to friction
E=0.300*36.5*9.81*4.30
E= 461.9 J
(c)
The work done by the normal force = 0 J as no displacement is taking place in that direction
(d)
The work done by the gravitational force 0 J, as there is zero displacement in that direction too.
(e)
the change in kinetic energy of the box
ΔKE=645 - 461.9 J
ΔKE= 183 J
(f)
The final speed of the box
0.5*36.5*v^2 = 183 J
v^2 = 10
v = 3.16 m/s
Answer:
(a) The work done by the applied force is [tex]645 J[/tex]
(b) The energy loss due to friction is [tex]461.9 J[/tex]
(c) The work done by the normal force is [tex]0 J[/tex]
(d) The work done by gravity is [tex]0 J[/tex]
(e) The change in kinetic energy of the box is [tex]183 J[/tex]
(f) The final speed of the box is [tex]3.16 m/s[/tex]
Explanation:
Given:
Weight of the box [tex]= 36.5 kg[/tex]
Distance [tex]= 4.30 m[/tex]
Horizontal force [tex]= 150 N[/tex]
Coefficient of friction between box and floor [tex]= 0.300[/tex]
Step 1:
(a) The work done is,
[tex]W=F\times d[/tex]
Substitute the values,
[tex]W=150\times 4.3[/tex]
[tex]W = 645 J[/tex]
Therefore, the work done by the applied force is 645 J.
Step 2:
(b) The increase in internal energy in the box-floor system due to friction is given by,
[tex]E=0.300\times 36.5\times 9.81\times 4.30[/tex]
[tex]E= 461.9 J[/tex]
Therefore, the energy loss due to friction is [tex]461.9 J.[/tex]
Step 3:
(c) The work done by the normal force is [tex]0 J.[/tex] Because, no displacement is taking place in that direction.
(d) The work done by gravity is also [tex]0 J[/tex], as there is zero displacement in that direction.
Step 4:
The change in kinetic energy of the box is given by,
Δ[tex]KE=645 - 461.9 J[/tex]
Δ[tex]KE= 183 J[/tex]
Therefore, the change in kinetic energy of the box is [tex]183 J.[/tex]
Step 5:
(f) The final speed of the box is given by,
[tex]0.5\times36.5\timesv^2 = 183 J[/tex]
[tex]v^2 = 10[/tex]
[tex]v = 3.16 m/s[/tex]
Therefore, the final speed of the box is [tex]3.16 m/s[/tex]
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