Part C What is the work Wr done by the tension force? Express your answers in joules to two significant figures. • View Available Hint(s) 1 of 1 Wr = -0.92 J Submit Previous Answers X Incorrect; Try Again; 4 attempts remaining Provide Feedback

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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Recall that the work WWW done by a constant force F⃗ F→F_vec at an angle θθtheta to the displacement Δr⃗ Δr→ is

W=F⃗ ⋅Δr⃗ =FΔrcosθW=F→⋅Δr→=FΔrcos⁡θ.

The vector magnitudes FFF and ΔrΔr are always positive, so the sign of WWW is determined entirely by the angle θθtheta between the force and the displacement.

What is the work WTWT done by the tension force?
Express your answers in joules to two significant figures.
I Review I Constants
Learning Goal:
To practice Tactics Box 9.1 Calculating the Work Done by a
Constant Force.
Part B
Recall that the work W done by a constant force F at an angle 0
to the displacement Ar is
What is the work Wn done on the box by the normal force?
Express your answers in joules to two significant figures.
W = F. Ar = FAr cos 0.
• View Available Hint(s)
The vector magnitudes F and Ar are always positive, so the
sign of W is determined entirely by the angle 0 between the force
and the displacement.
Wn = 0 J
Submit
Previous Answers
Correct
Because there is no component of the normal force in the direction of motion, the normal force does no work on the box.
Part C
What is the work WT done by the tension force?
Express your answers in joules to two significant figures.
• View Available Hint(s)
Figure
1 of 1
ΑΣΦ
?
WT =
- 0.92
J
47
Submit
Previous Answers
X Incorrect; Try Again; 4 attempts remaining
Provide Feedback
Next >
Transcribed Image Text:I Review I Constants Learning Goal: To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. Part B Recall that the work W done by a constant force F at an angle 0 to the displacement Ar is What is the work Wn done on the box by the normal force? Express your answers in joules to two significant figures. W = F. Ar = FAr cos 0. • View Available Hint(s) The vector magnitudes F and Ar are always positive, so the sign of W is determined entirely by the angle 0 between the force and the displacement. Wn = 0 J Submit Previous Answers Correct Because there is no component of the normal force in the direction of motion, the normal force does no work on the box. Part C What is the work WT done by the tension force? Express your answers in joules to two significant figures. • View Available Hint(s) Figure 1 of 1 ΑΣΦ ? WT = - 0.92 J 47 Submit Previous Answers X Incorrect; Try Again; 4 attempts remaining Provide Feedback Next >
II Review l Constants
Learning Goal:
To practice Tactics Box 9.1 Calculating the Work Done by a
Constant Force.
TACTICS BOX 9.1 Calculating the work done by a constant force
Sign of
W
Force and displacement
Work W
Energy transfer
Recall that the work W done by a constant force F at an angle 0
to the displacement Ar is
W = F· AF = FAr cos 0.
AF
0°
F(Ar)
+
The vector magnitudes F and Ar are always positive, so the
sign of W is determined entirely by the angle 0 between the force
and the displacement.
Energy is transferred into the system. The particle speeds up. K
increases.
AF
< 90° | F(Ar) cos 0
+
90°
No energy is transferred. Speed and K are constant.
AF
> 90° F(Ar) cos 0
Energy is transferred out of the system. The particle slows down. K
decreases.
Figure
1 of 1
AF
180°
-F(Ar)
A box with weight of magnitude FG
in (Figure 1) at left. The normal force acting on the box has a magnitude n =
down the inclined plane.
2.00 N is lowered by a rope down a smooth plane that is inclined at an angle o = 30.0 ° above the horizontal, as shown
1.73 N, the tension force is 1.00 N, and the displacement A7 of the box is 1.80 m
Part A
平
Transcribed Image Text:II Review l Constants Learning Goal: To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. TACTICS BOX 9.1 Calculating the work done by a constant force Sign of W Force and displacement Work W Energy transfer Recall that the work W done by a constant force F at an angle 0 to the displacement Ar is W = F· AF = FAr cos 0. AF 0° F(Ar) + The vector magnitudes F and Ar are always positive, so the sign of W is determined entirely by the angle 0 between the force and the displacement. Energy is transferred into the system. The particle speeds up. K increases. AF < 90° | F(Ar) cos 0 + 90° No energy is transferred. Speed and K are constant. AF > 90° F(Ar) cos 0 Energy is transferred out of the system. The particle slows down. K decreases. Figure 1 of 1 AF 180° -F(Ar) A box with weight of magnitude FG in (Figure 1) at left. The normal force acting on the box has a magnitude n = down the inclined plane. 2.00 N is lowered by a rope down a smooth plane that is inclined at an angle o = 30.0 ° above the horizontal, as shown 1.73 N, the tension force is 1.00 N, and the displacement A7 of the box is 1.80 m Part A 平
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