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vtstudent
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1. An 8.3 kg crate is pulled 4.8 m up a 30 degree incline by a rope angled 18 degrees above the incline. The tension in the rope is 120 N and the crate's coefficient of kinetic friction on the incline is 0.24. How much work is done by tension, by gravity, and by the normal force? What is the increase in thermal energy of the crate and incline?
2. W=F[tex]_{}x[/tex]d[tex]_{}s[/tex]
F[tex]_{}g[/tex]=mg
F[tex]_{}fr[/tex]=uN
F[tex]_{}N[/tex]=mgsin30
I think the work done by the normal force is 0 because it is perpendicular to the motion of the crate.
For the work done by gravity, I found mg = 8.3*9.8=81.34
Then, for the component of gravity parallel to the motion of the crate, I found 81.34sin30=40.67. I multiplied this by distance to get 40.67*4.8=195.216 as the work done by gravity
For work done by the tension force, I found the component of tension parallel to the motion of the crate. 120cos18=114.126
Then, I multiplied this by distance to get 547.81 as the work done by the tension force.
Now, to deal with friction I am a bit lost. I found the normal force. Normal force = mgcos30 = 8.3(9.8)cos30=70.44
Then I found the friction force. F(friction)=.24(70.44)=16.91N
Do I multiply this by distance (16.91*4.8=81.15N) and then add it to the work done by the tension force, because the tension force must overcome the friction force as well?
2. W=F[tex]_{}x[/tex]d[tex]_{}s[/tex]
F[tex]_{}g[/tex]=mg
F[tex]_{}fr[/tex]=uN
F[tex]_{}N[/tex]=mgsin30
The Attempt at a Solution
I think the work done by the normal force is 0 because it is perpendicular to the motion of the crate.
For the work done by gravity, I found mg = 8.3*9.8=81.34
Then, for the component of gravity parallel to the motion of the crate, I found 81.34sin30=40.67. I multiplied this by distance to get 40.67*4.8=195.216 as the work done by gravity
For work done by the tension force, I found the component of tension parallel to the motion of the crate. 120cos18=114.126
Then, I multiplied this by distance to get 547.81 as the work done by the tension force.
Now, to deal with friction I am a bit lost. I found the normal force. Normal force = mgcos30 = 8.3(9.8)cos30=70.44
Then I found the friction force. F(friction)=.24(70.44)=16.91N
Do I multiply this by distance (16.91*4.8=81.15N) and then add it to the work done by the tension force, because the tension force must overcome the friction force as well?