03work and energy

f = µ k N = µ k mg
B O A
B
v v v d W = F ⋅ d r = − F dr
v W = ∫ − F d r = ∫ − Fds
A B
v dr
v f
= − µ k mg ∫ ds = − µ k mgs = − µ k mg πR
A
B
AHale Waihona Puke (b) If it moves along the straight line, The work depends on the path .
v v v v 1 ( f + p + N ) ⋅ s = 0 − mv 2 2 v v N ⋅s = 0 N = mg cosθ
v s
v v0
π v v p ⋅ s = mgs cos( + θ ) 2 f =µk mg cosθ
v v f ⋅ s = − µ k mg cosθ s
v v N fk
r r r db da r d r r + ⋅b (a ⋅ b ) = a ⋅ dt r dt dt r r r r db d a d r + ×b (a × b ) = a × dt dt dt
r r r r a × b = −b × a
Example Calculate the work a person does to stretch a spring slowly from the normal length xa=0 to an extra length xb= x .
0 ≤θ <
π
2
,W > 0
π
2
< θ ≤ π ,W < 0
θ =
π
2
,W = 0
Where θ is the angle between the force and the displacement of its point of application.
Example :The force exerted on an object is
= 3 × ( 3 − 1) + 4 × (5 − 2) = 18J
7-2 Work done by a varying force • The work done for an infinitesimal displacement
b
v dr
(L)
v v dW = F ⋅ dr
• The total work during the the displacement from a to b is
θ v F
a
v v A = ∫ F ⋅ dr =
a( L)
b
a( L)
v ∫ F dr cosθ = ∫ F cosθ ds
a( L)
b
b
This is the most general definition of work.
In rectangular coordinates,
v ˆ+F ˆ ˆ F = Fx i j F k + y z
dp dv d F= =m = m ( 3t 2 + 2t ) = m ( 6t + 2) dt dt dt
Calculate the total work .
v v W = ∫ F ⋅ dr =
0
dx Fdx = ∫ F dt = ∫ dt T = ∫ m ( 6 t + 2 )( 3 t 2 + 2 t )d t
v v Wab = ∫ F ⋅ dr =
a( L)
b
a( L)
v v v v ∫ ( F1 + F2 + L + Fn ) ⋅ dr
b
v v = ∫ F1 ⋅ dr +
a( L)
b
a( L)
v v ∫ F2 ⋅ dr + L +
b
a( L)
v v ∫ Fn ⋅ dr
b
= W1ab + W2ab + L + Wnab
1 K = mv 2 2
−1
− 1
Relativistic kinetic energy
2. Work-energy principle The net work done on an object equals to the change in its kinetic energy.
v v W = ∫ F ⋅ dr =
a( L) b b
v ˆ + d yˆ ˆ dr = d xi j + d zk
a ( L)
∫ (F dx + F dy + F dz )
x y z
•The total work of several forces Consider a particle being from a to b along path L v v displaced v under the action of F1 , F2 L , Fn v The net force is F
∆ W dW P = lim = ∆t → 0 ∆ t dt v v v v dW F ⋅ dr P= = = F ⋅ V = FV cos θ dt dt
The SI unit of power = Watt (W), where 1 W= 1 J per sec=1 kg.m2/s3 Note that the imperial unit of horsepower (hp) is still used, for example for cars. 1hp = 746 W The amount of work done is sometimes expressed as the product of the power output multiplied by time taken for this. A common unit for this is the kilowatt-hour, where 1kWh = 1000 × 3600 J = 3.6 ×106J = 3.6MJ.
T
∫ Fv d t
= 2 m ∫ ( 9 t 3 + 9 t 2 + 2 t )d t = m ( 9T 4 + 12 T 3 + 4T 2 ) 2 0
Example An object of mass m is displaced between two points A and B on a table. The coefficient of kinetic friction between the book and table top is µk . Calculate the work done by the friction exerted on the object by the table top if the object moves along (a)the semicircular path as shown in the figure ;(b)the straight line. Solution (a)
A
B
1 1 2 2 = mv B − mv A = KB − KA 2 2
Example An object moves up along the plane inclined at θ with the horizontal from its bottom with an initial speed v0, as shown in the figure. Later, it slides down the plane. The object has the speed vf when it arrives at the bottom of the plane again. Determine the maximum height reached by the object. Solution： It moves up with initial speed v0 and stops at the height h. Apply work –kinetic energy theorem.
Scalar product and vector product
r r r r a⋅b = b⋅a
r r r r 2 a ×a = 0 a⋅a = a r r r r r r r r r c × (a × b ) = a (b ⋅ c ) − b (a ⋅ c ) r r r r r r r r r c ⋅ (a × b ) = a ⋅ (b × c ) = b ⋅ (c × a )
v v v F = 3i + 4 j ( N )
Find the work done in moving the object from(1,2) (m)to (3,5) (m). Solution:
y 6 4 2 1 3 5 x
v r W = F ⋅ ∆r
= Fx ⋅ ∆ x + F y ⋅ ∆ y
W AB
1 1 2 2 = ∆K = K B − K A = mv B − mv A 2 2
(A → B) B
v v v dv v dW = F ⋅ dr = Fr dr = m dr
v dr = m dv dt
W AB = ∫ dW =
A B
dt
v Fr
A
v dr
= mv dv
v F
∫ mvdv
The total work done is just the scalar addition of the work done by each force.