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:
AP Physics C
W = Frcosq
When the work done by an external force onto the system is:
+ : the system gains energy
- : the system loses energy
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
then the work done by that force is zero.
What force(s) acting on the sled does no 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)
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.
W = FDr cos q
W = F(x) dx
W = F • dr
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
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)
Force exerted by spring: Fs = - kx
An archer pulls her bowstring back 0.4m by exerting a force that increases uniformly from 0 to 230N.
Ans: a) 575N/m, b) 46.0J