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: AxB = AB sin q Vector

W= F·r Work is a dot product Work is the dot product of Force and displacement. W= F·r

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)

When the work done by an external force onto the system is: 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 180 o = -1 => W= F r and work is negative Which force acting on the sled does 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?

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 Fxnet to calculate N) (A: -136J)

Work 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 is the area under an F vs d graph.

Work Done by a Variable Force For a small displacement, Dx, W ~ F Dx Over the total displacement:

Work Done by a Variable Force The work done is equal to the area under the curve.

Work W = F Dr cos q W =  F(x) dx W =  F • dr

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.

Work Done by a Spring

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