You push your physics book a distance 1.41 m along a horizontal tabletop with a horizontal force of 2.34 N. The opposing force of friction is 0.550 N.
How much work does your force 2.34 N do on the book?
What is the work done on the book by the friction force?
What is the total work done on the book?
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2006-06-28 15:35:29 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Work done by the force 2.34 N is, ‘force x distance through which the force is moved’.
Therefore, it is = 2.34 N x 1.41m = 3.2994 joule.
Work done by the frictional force 0.55 N is, ‘force x distance through which the force is moved’.
Therefore, it is = 0.55 N x 1.41m = 0.7755 joule.
The last question is vague.
The total work is the work done by the force 2.34 N and it is equal to 3.2994 joule.
Out of this total work a part (3.2994 -0.7755 = 2.5239) joule is used to accelerate the book and 0.7755 joule is dissipated as heat energy.
2006-06-28 19:27:02 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
For the first part (pushing the book): W = Fd (Work = Force à distance). This would be W = 2.34N à 1.41m, which gives us a result of 3.2994N-m.
For the second part (friction resisting the push): Wf = fd (friction Work à friction distance). This would be Wf = 0.55N à 1.41m, which gives us a result of 0.7755N-m.
Moving on to the third part, the total work performed upon the book, we can use Wt = W - Wf. This would be 3.2994N-m - 0.7755N-m, which gives us a result of 2.5239N-m (joules) total work done on the book.
2006-06-29 00:17:56 · answer #2 · answered by Dragosani 3 · 0⤊ 0⤋
what?????
2006-06-28 23:06:33 · answer #3 · answered by minniemouse493 3 · 0⤊ 0⤋