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Finding the power and work needed

  • #1
This is the orginal problem.....a.)How much power is needed to push a 95 kg chest at 0.62 m/s along a horizontal floor where the coefficient of friction is 0.78.(ans. 450W)... b.) How much work is done in pushing the chest 11 m? (ans. 7.99x10^3 J)

New question: Answer the previous question when the floor slopes upward at an angle of 5 degrees, and you push parallel to the floor.



Fa=(coefficient of friction)(mg)
P=(F) (v)
W=(F) (delta x)



My attempt is W=(95kg)(9.81m/s^2)(11m)(sin5 +cos5(.78))= 8.81 x10^3 J so solving for power I would use Work/17.74= 496 Watts
.
A different route is P=(.78)(95kg)(9.8m/s^2)(cos5)(.62)=448 W
(solving for t I get 11m/.62m/s=17.74 s) so W=448x17.74 which is 7.96 x10^3 J
-both of these answers are smaller so I do not believe this is true
I am unsure if either of these is right and if both are wrong where have I gone wrong! thank you
 

Answers and Replies

  • #2
PhanthomJay
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This is the orginal problem.....a.)How much power is needed to push a 95 kg chest at 0.62 m/s along a horizontal floor where the coefficient of friction is 0.78.(ans. 450W)... b.) How much work is done in pushing the chest 11 m? (ans. 7.99x10^3 J)

New question: Answer the previous question when the floor slopes upward at an angle of 5 degrees, and you push parallel to the floor.



Fa=(coefficient of friction)(mg)
P=(F) (v)
W=(F) (delta x)



My attempt is W=(95kg)(9.81m/s^2)(11m)(sin5 +cos5(.78))= 8.81 x10^3 J so solving for power I would use Work/17.74= 496 Watts
this looks good
.
A different route is P=(.78)(95kg)(9.8m/s^2)(cos5)(.62)=448 W
what happened to the sin5 piece... ?? The push force and power required must overcome both friction and gravity.
 
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