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Work and transformation of energy

  • Thread starter pharm89
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Homework Statement


Hi I am taking a grade 11 physics correspondance course and would like to know if iam on the right track. thanks in advance.

1. (a) A force of 280 N is required to push a box 8.2 m across a rough floor. How much work is done?

(b) Explain 1 transformation of energy that occured in part a.


(e) It is found that 1.50 X 10 ^3 J of work are required to push a box across a floor a horizontal distance of 8.0 m . What average force is being exerted on the box?

Homework Equations


W = F X D



The Attempt at a Solution



(a) W= F X D
=280 N X 8.2 m
=2296 J
=2.2 X 10 ^3

(b) A force is provding energy that will be everntually transformed into heat energy through friciton between the force and the rough floor. The force is acting against the force of friciton.

(c) there is no work being done. This is becuase conditions for work are quite specfic and in this problem a displacement occurs but the force being exerted does not cause the displacement.

(d) W = F X D cos (theta)
=104 n X 5.0 m X cos 32 degrees
=440.99 J
=441 J

(e) F = W/ D
=150 J / 8.0 m = 18.75 N
Thanks
Pharm 89

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
G01
Homework Helper
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Just so you know, there are answers you posted for questions that you didn't provide, so I can't really help you with those. But you seem to know what your doing otherwise. Nice Job!
 
  • #3
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Just so you know, there are answers you posted for questions that you didn't provide, so I can't really help you with those. But you seem to know what your doing otherwise. Nice Job!
whoops question c and d

(c) A 46 kg woman climbs a flight of stairs, 6.2 m high. How much work does she do.

(d) A force of 104 N is applied to a 26 kg mass at an angle of 32 degrees to the horizontal. Calculate the work done when the mass is moved through a horizontal distance of 5.0m .

Thanks
Pharm89
 
  • #4
PhanthomJay
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whoops question c and d

(c) A 46 kg woman climbs a flight of stairs, 6.2 m high. How much work does she do.

(d) A force of 104 N is applied to a 26 kg mass at an angle of 32 degrees to the horizontal. Calculate the work done when the mass is moved through a horizontal distance of 5.0m .

Thanks
Pharm89
For problem (c), what is the change in the person's energy at the top of the stairs, and how would you account for that change?
 
  • #5
HallsofIvy
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c. To lift an object, even yourself, you must exert an upward force equal to the the object's weight. Here the woman must exert an upward force of 46kg* g= 46*9.81 Newtons for an upward distance of 6.2 m.

PhantomJay is suggesting looking at the change in potential energy but you may not have had that.

The other problems look good.
 
  • #6
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c. To lift an object, even yourself, you must exert an upward force equal to the the object's weight. Here the woman must exert an upward force of 46kg* g= 46*9.81 Newtons for an upward distance of 6.2 m.

PhantomJay is suggesting looking at the change in potential energy but you may not have had that.

The other problems look good.
Thanks for the help.

therefore the conditions for work are met.
g= 46kg X 9.8 N
=450.8 N
W=450.8 N X 6.2 m
W=2794.96 J = 2795 J
 

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