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Work. AP Physics C Mrs. Coyle. Work. The work done on an object by an external constant force: W = F r cos q  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:

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Work

AP Physics C

Mrs. Coyle


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
Products of two Vectors A and B

  • Dot Product:

    • A·B =A B cos q Scalar

  • Cross Product:

    • AxB = AB sin q Vector


Work is a dot product
Work is a dot product

  • Work is the dot product of Force and displacement.

    W= F·r


Units of work
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
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


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
Question

  • Does the centripetal force do work?


Example 1

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 1


Work done by resultant
Work done by Resultant move the 6kg box along the floor by 20m?

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 is the area under an f vs d graph
Work is the area under move the 6kg box along the floor by 20m? an F vs d graph.


Work done by a variable force
Work Done by a move the 6kg box along the floor by 20m? Variable Force

  • For a small displacement, Dx, W ~ FDx

  • Over the total displacement:


Work done by a variable force1
Work Done by a Variable Force move the 6kg box along the floor by 20m?

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


Work move the 6kg box along the floor by 20m?

W = FDr cos q

W =  F(x) dx

W = F • dr


Example 2 work by a variable force
Example 2 – Work by a Variable Force move the 6kg box along the floor by 20m?

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
Example 3 Work as dot product of vectors move the 6kg box along the floor by 20m?

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
Hooke’s Law move the 6kg box along the floor by 20m?

Force exerted by spring: Fs = - kx

  • equilibrium position is x = 0

  • k : spring constant (force constant) (k increases as stiffness of the spring increases)


F s is a r estoring force
F move the 6kg box along the floor by 20m? s 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.


Work done by a spring
Work Done by a Spring move the 6kg box along the floor by 20m?


Example 16
Example #16 move the 6kg box along the floor by 20m?

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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