Summary
Summary Gr 11 Physics Nelson Ch.5 Notes
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- Physics
- Institution
- 11th Grade
It summarizes all the main points according to the Nelson Grade 11 textbook, with diagrams provided.
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Chapt& 5 W*k, En&gy, Pow&, and Society5.1 Work
Mechanical work (W)
➲ applying a force on an object that displaces the object in the direc=on of the force or a component of
the force
Work Done by a Constant Force
The mechanical work, W, done by a force on an object is the product of the magnitude of the force, F,
and the magnitude of the displacement of the object, ∆𝑑:
𝑾 = 𝑭∆𝒅
This equa=on applies only to cases where the magnitude of the force is constant and where the force
and the displacement are in the same direc=on.
Since work is a product of the magnitude of the force and the magnitude of the displacement, the
symbols for force and displacement in the equa=on are wriHen without the usual vector nota=ons 𝐹⃗ , ∆𝑑⃗.
Work is a scalar quantity; there is no direction associated with work.
• Work Done When Force and Displacement Are in the Same Direc=on (Page 222 -223)
The equa=on 𝑊 = 𝐹∆𝑑 may be used to calculate the amount of mechanical work
done on an object when force and displacement are in the same direction.
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,• Work Done When Force and Displacement Are in Different Direc=ons (Page 223-224)
In some cases, an object may experience a force in one direc=on while the object moves in
a different direc=on.
For example, this occurs when:
A person pulls on a suitcase with wheels and a handle (Figure a).
The free-body diagram (FBD) for this situa=on is shown in (Figure b).
In this case, the applied force is directed toward the person’s hips, while the suitcase rolls
on the floor in the forward direc=on.
The FBD shows all the forces ac=ng on the suitcase, including the force of gravity (F -⃗! ),
-⃗" ) and the applied force (F
the normal force (F -⃗# ) resolved into horizontal (F$ ) and
ver=cal (F% ) components. (We’ll assume that the force of fric=on is negligible)
The FBD also indicates that the applied force vector -F⃗# , makes an angle 𝜃 with the
displacement vector, , ∆𝑑⃗.
The magnitude of F$ is given by the equation F$ = F# cos 𝜃
Since the horizontal component, F$ , is the only force in the direction of the suitcase’s
displacement, it is the only force that causes the suitcase to move along the floor. Thus,
the amount of mechanical work done by F$ on the suitcase may be calculated using the
equation:
𝑾 = 𝑭 (𝒄𝒐𝒔 𝜽) ∆𝒅
o W: work done
o F: force
o cos 𝜃: angle between the force and the displacement vector ∆d
2|Page
, F% does no work on the suitcase because the suitcase is not displaced in the direction of F%
Since F% is perpendicular to the suitcase’s displacement along the floor (cos 𝜃 = 90°), F%
does not work on the suitcase.
The above example can be calculated using the work equation 𝑾 = 𝑭 (𝒄𝒐𝒔 𝜽) ∆𝒅, where
𝜃 = 90°:
𝑾 = 𝐅⃗𝐚 (𝒄𝒐𝒔 𝜽) ∆𝒅
= 𝐅⃗𝐚 (𝒄𝒐𝒔 𝟗𝟎°) ∆𝒅
= 𝐅⃗𝐚 (𝟎)∆𝒅, since 𝒄𝒐𝒔 𝟗𝟎° = 𝟎𝟎
= 𝟎𝐉
This result illustrates an important principle of mechanical work.
In general, the work done by a force is zero when the force’s direction is perpendicular to
the object’s displacement; that is when 𝜃 = 90°, cos 90° = 0, and when W = 0 J.
• Work Done When a Force Fails to Displace an Object (Page 225-226)
In some cases, a force is applied on an object, but the object does not move: no
displacement occurs.
For example, when you stand on a solid floor, your body applies a force on the floor equal
to your height ( -F⃗! = -F⃗# ), but the floor does not move. Your body does no work on the
floor because the floor is not displaced.
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