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Work

- The work done on an object by an external constant force:

W = Frcosq

- is the angle between the displacement, r, and the force F.

Products of two Vectors A and B

- Dot Product:
- A·B =A B cos q Scalar
- Cross Product:
- AxB = AB sin q Vector

Units of Work

- Is a scalar quantity.
- Unit: Joule 1 J = 1 N∙m
- James Prescott Joule, English, 19thCentury
- Other units of work:
- British system:
- foot-pound
- cgs system:
- erg=dyne ·cm
- Sub-atomic level:
- electron-Volt (eV)

Meaning of Sign of Work

When the work done by an external force onto the system is:

+ : the system gains energy

- : the system loses energy

When the force is in the direction of the

displacement, =0o:

cos 0= 1 => W= F r and work is positive

Which force acting on the sled does positive work?

When the force acts opposite to the direction of

the displacement, =180o:

cos 180o = -1 => W= F r and work is negative

Which force acting on the sleddoes negative work?

When the force acts perpendicular to the

displacement,

then the work done by that force is zero.

What force(s) acting on the sled does no work?

Question

- Does the centripetal force do work?

How much work was done by the 30N applied tension force to move the 6kg box along the floor by 20m? (A: 300J)

How much work did the weight do in the same case?

If µ=0.2, how much work did the friction do? (Hint: Use Fxnetto calculate N)(A: -136J)

Example 1Work done by Resultant

W= W1+ W2+ Wn

Since work is a scalar, the net work done by

a resultant force is equal to the sum of the

individual works done by each individual

force acting on the object.

Work Done by a Variable Force

- For a small displacement, Dx, W ~ FDx
- Over the total displacement:

Work Done by a Variable Force

- The work done is equal to the area under the curve.

Example 2 – Work by a Variable Force

The force acting on a particle is F(x)= (9x2 +3) N.

Find the work done by the force on the particle as it moves it from x1 =1.5m to x 2=4m.

Ans: 199 J

Example 3 Work as dot product of vectors

A particle is pushed by a constant force of

F=(4i +5j)N and undergoes a displacement of

Dr= (3i+ 2j)m.

Find the work done by the force.

Ans: (12 +10)J= 22J (Scalar Quantity)

Hooke’s Law

Force exerted by spring: Fs = - kx

- equilibrium position is x = 0
- k : spring constant (force constant) (k increases as stiffness of the spring increases)

Fs is a Restoring Force

- because it always acts to bring the object towards equilibrium
- the minus sign in Hooke’s law indicates that the Fs always acts opposite the displacement form equilibrium.

Example #16

An archer pulls her bowstring back 0.4m by exerting a force that increases uniformly from 0 to 230N.

- What is the equivalent spring constant of the bow?
- How much work does the archer do in pulling the bow?

Ans: a) 575N/m, b) 46.0J

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