Chapter

Work and Kinetic Energy

Blinn College - Physics 1401 - Terry Honan

. - Introduction to Work

Mechanical Advantage

In Chapter D we considered the example of a pulley system lifting a weight.  We saw that using multiple pulleys one can lift a heavy object with a smaller force.  In the example, a weight W could be lifted by a tension T=W/5, but the smaller force must act over a larger distance.  To lift the weight by Δ y one must pull on the rope by Δ x=5 Δ y.  This suggests that force times distance is an important quantity; we will define it as the work.

One Dimensional Work by a Constant Force

If in one dimension we move something by a displacement Δ x with a constant force F.  We will define the work done in this case by

W=F Δ x.

Note that even in one dimension force and displacement are vector quantities; a one dimensional vector is a real number and the sign gives its direction.  It follows that if the force and displacement are in the same direction then the work is positive and if opposite it is negative.  The work is a scalar quantity.

Units:  The SI unit for work and energy is:  (QuantityUnit)=joule=(QuantityUnit)(QuantityUnit)

Two and Three Dimensional Work by a Constant Force

We now need to generalize this to the case of a constant force (OverVector[F]) and a straight-line path in two or three dimensions. Here we have a displacement  (OverVector[d]) and we will define θ to be the angle between the two vectors  (OverVector[F]) and  (OverVector[d]).  What is important is the component of the force along the direction of the displacement, which will be written (Subscript[F, ∥]) and is given by (Subscript[F, ∥])=F cos θ.

The definition of work becomes

W=(Subscript[F, ∥]) d=F d cos θ=F (Subscript[d, ∥])

where d=‖ (OverVector[d])‖ is the magnitude of the displacement  (OverVector[d]) and (Subscript[d, ∥]) is the component of  (OverVector[d]) parallel to the force (OverVector[F]).

Example . - Dragging a Crate

Solution

T=115 (QuantityUnit), f=80 (QuantityUnit), m=38 (QuantityUnit), θ=35°  and  d=4 (QuantityUnit)

W=F d cos θ.

(Subscript[W, T])=T d cos θ=377.(QuantityUnit)

(Subscript[W, f])=f d cos 180°=-f d=-320(QuantityUnit)

Solution

(Subscript[F, net,hor])=T cos θ-f=m a  ⟹  a=(1/m)(T cos θ-f)=0.374(((QuantityUnit))/(((QuantityUnit))^2))

(v^2)=(Subscript[v, 0]^2)+2 a Δ x   ⟹   (v^2)=(Subscript[v, 0]^2)+2 a d   ⟹   v=(Sqrt[2 a d ])=1.73(QuantityUnit)/(QuantityUnit)

. - Work in General

So far we have the definition of work for a constant force and a straight line path to be W=(Subscript[F, ∥]) d=F d cos θ=F (Subscript[d, ∥]).  We need to generalize this to a varying force acting on a general path that is allowed to curve. Break up the path into many small segments  ((OverVector[d])^((i))); the force at that small segment ((OverVector[F])^((i))) will be essentially constant if the segment is sufficiently small. The work over that small segment is then (Subscript[F, ∥]^((i)))(d^((i))) The generalized definition of work will then be summing the work over all the small segments

W=(∑_i)(Subscript[F, ∥]^((i)))(d^((i)))    (small ((OverVector[d])^((i))))

This expression will not be used in calculations this semester. It will be used for definitions.

Constant Force

If a force is constant then (OverVector[F])=((OverVector[F])^((i))) and them we can take it out of the above sum.

W=(∑_i)(Subscript[F, ∥]^((i)))(d^((i)))=(∑_i)F (Subscript[d, ∥]^((i)))=F(∑_i)(Subscript[d, ∥]^((i))) =F (Subscript[d, ∥])=(Subscript[F, ∥]) d=F d cos θ     (small ((OverVector[d])^((i))))

Here (OverVector[d]) is the displacement vector from the start of the path to its end: (OverVector[d])=(∑_( i))((OverVector[d])^((i))).

This means the the work for constant force is the same as the work for a constant force over the straight-line path, where (OverVector[d]) is the displacement vector from the start to the end of the path.

Work Done by Gravity

An important special case of the previous result is the work done by gravity. The force (OverVector[F])=-m g  (OverHat[y]) is a constant and the component of a displacement (OverVector[d]) in the direction of gravity is: (Subscript[d, ∥])=-Δ y.

(Subscript[W, grav])=F (Subscript[d, ∥])=-m g Δ y.

Example . - A Box on a Table

Solution

(Subscript[W, grav])=-m g Δ y=m g h.

Solution

(Subscript[W, grav])=-m g Δ y=-m g h.

Solution

(Subscript[W, grav])=-m g Δ y=0.

One Dimensional Work by a Varying Force

If F(x) is a one dimension force as a function of position x, the work when moving from position (Subscript[x, i]) to (Subscript[x, f]) is the area under the F vs. x graph between (Subscript[x, i]) and (Subscript[x, f]). The convention when discussing area under a curve is the when F(x) Δ x is negative, the contribution to the area is negative.

Note that when (Subscript[x, f])<(Subscript[x, i]), then Δ x<0 and the sign of F(x) Δ x is reversed; positive F then gives a negative area and negative F gives a positive area.

Example . - Work Done by a Varying Force

Solution

(Subscript[W, 6->30])=(Subscript[W, 6->10])+(Subscript[W, 10->12])+(Subscript[W, 12->18])+(Subscript[W, 18->30])

(Subscript[W, 6->10])=(1/2)(4(QuantityUnit)×72(QuantityUnit))=144(QuantityUnit)

(Subscript[W, 10->12])=2(QuantityUnit)×72(QuantityUnit)=144(QuantityUnit)

(Subscript[W, 12->18])=(1/2)(6(QuantityUnit)×72(QuantityUnit))=1216(QuantityUnit)

(Subscript[W, 18->30])=-(1/2)(12(QuantityUnit)×120(QuantityUnit))=-864(QuantityUnit)

(Subscript[W, 6->30])=(Subscript[W, 6->10])+(Subscript[W, 10->12])+(Subscript[W, 12->18])+(Subscript[W, 18->30])=-360(QuantityUnit)

Work Done by a Spring

In chapter 6, we discussed Hooke’s law, the force law of a spring. The force of a spring acting on something is F=-k x. We may now obtain an expression for the work done by a spring. Apply the preceding discussion of the work done in one dimension by a varying force F(x) using the Hooke's law  F(x)=-k x.

The negative area between (Subscript[x, i]) and (Subscript[x, f]) is just the negative area of the triangle from the origin to (Subscript[x, f]) to F((Subscript[x, f])) subtracting the missing part, the negative area of the triangle from he origin to (Subscript[x, i]) to F((Subscript[x, i]))

(Subscript[W, spring])=(1/2)(Subscript[x, f]) F((Subscript[x, f]))-(1/2)(Subscript[x, i]) F((Subscript[x, i]))=(1/2)(Subscript[x, f]) (-k (Subscript[x, f]))-(1/2)(Subscript[x, i]) (-k (Subscript[x, i]))

This gives the expression.

(Subscript[W, spring])=-(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2)).

Example . - Hooke’s Law

Solution

F=60(QuantityUnit)   and   x=0.04(QuantityUnit)

F=k x   ⟹   k=F/x=1500(QuantityUnit)/(QuantityUnit).

Solution

(Subscript[W, spring])=-(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2))=-(1/2)k ((x^2)-(0^2))=-(1/2)k (x^2)=-1.2(QuantityUnit) .

. - Power

Power, in the most general sense, is the rate that something uses or provides energy.

(Subscript[𝒫, ave])=(Energy/(Δ t)).

The power delivered by a motor or engine is the rate that it can do work

(Subscript[𝒫, ave])=(W/(Δ t)).

For the one-dimensional case we can write the work in terms of the force and velocity. For a small d=Δ x, since v=Δ x/Δ t we get:

(Subscript[𝒫, ave])=(W/(Δ t))=((F Δ x)/(Δ t))=F v

Units:  The SI unit for power is:  (QuantityUnit)=watt=(QuantityUnit)/(QuantityUnit)

. - Kinetic Energy and the Work-Energy Theorem

The Net Work

Newton's second law isn't a general statement about forces but is about the net force acting on a body,  (Subscript[(OverVector[F]), net])=m (OverVector[a]).  (Subscript[(OverVector[F]), net]) is the vector sum of all forces acting on a body; we use a free-body diagram to help us sum these forces.  Let us symbolically write the net force in terms of all the forces acting on a body (all the forces in the free-body diagram) as

(Subscript[(OverVector[F]), net])=(Subscript[(OverVector[F]), 1])+(Subscript[(OverVector[F]), 2])+....

Each force acting on a body does work on the body.  (Some of these works may be zero, however.)  If the work done by (Subscript[(OverVector[F]), i]) is labeled (Subscript[W, i]) then we can define the net work as the sum of all these works.

(Subscript[W, net])=(Subscript[W, 1])+(Subscript[W, 2])+....

The Work-Energy Theorem

The work-energy theorem is a very important result.  It is where the idea of energy comes into physics and it explains why work is a useful notion.  We define the kinetic energy by

K=(1/2)m (v^2).

The theorem is quite simple to state; it equates the net work to the change in the kinetic energy.

(Subscript[W, net])=Δ K

where the change is kinetic energy is  Δ K=(Subscript[K, f])-(Subscript[K, i])=(1/2)m  ((Subscript[v, f]^2) - (Subscript[v, i]^2)).

Example . - Dragging a Crate (continued)

(Subscript[W, T])=376.8(QuantityUnit) ,  (Subscript[W, f])=-320.0(QuantityUnit)  and (Subscript[W, N])=0=(Subscript[W, grav])

Solution

(Subscript[W, net])=(Subscript[W, T])+(Subscript[W, f])+(Subscript[W, N])+(Subscript[W, grav])=376.8(QuantityUnit)-320.0(QuantityUnit)+0+0=56.8(QuantityUnit)

Solution

(Subscript[W, net])=Δ K=(1/2)m ((Subscript[v, f]^2)-(Subscript[v, i]^2))

(Subscript[v, f])=(Sqrt[(2/m)(Subscript[W, net])])=1.73(QuantityUnit)/(QuantityUnit)

Understanding the Work-energy Theorem

Consider a single mass moving along a general path. The net work is the sum of the works done by each force acting on a body. Since the sum of all the forces is the net force, the net work becomes the work done by the net force. We can understand this theorem using Newton’s second law and considering a small displacement (OverVector[d]).

(Subscript[W, net])=(Subscript[F, net,∥])d=m (Subscript[a, ∥])d=(m/2) ×2(Subscript[a, ∥])d

Choose coordinates so that this small displacement is in the x-direction, then we can write d=Δ x and (Subscript[a, ∥])=(Subscript[a, x]). For a small displacement, the force is constant and then so is the acceleration. The kinematics equations for constant acceleration gives

2(Subscript[a, ∥])d=2 (Subscript[a, x])Δ x=(Subscript[v, x]^2)-(Subscript[v, 0,x]^2)=(Subscript[v, f,x]^2)-(Subscript[v, i,x]^2)=(Subscript[v, f]^2)-(Subscript[v, i]^2)

The reason for the final equality in the expression is that at that instant the particle is moving in the x-direction only. Inserting this into the previous result gives the Work-Energy theorem, at least for a small displacement.

(Subscript[W, net])=(m/2)((Subscript[v, f]^2)-(Subscript[v, i]^2))=Δ K

For motion along a general path, the net work will be the sum over the net work of each small segment making the path. Adding all the Δ K for each small displacement gives the total Δ K for the entire path. The work-energy theorem follows.

(Subscript[W, net])=Δ K

Example . - Work Done by a Varying Force (continued)

Solution

m=30.0(QuantityUnit)  and  (Subscript[v, i])=v(6(QuantityUnit))=7.00(QuantityUnit)/(QuantityUnit)

(Subscript[W, net])=(Subscript[W, 6->30])=-360(QuantityUnit)

(Subscript[W, net])=(1/2)m ((Subscript[v, f]^2)-(Subscript[v, i]^2))   ⟹   (Subscript[v, f]^2)-(Subscript[v, i]^2)=((2 (Subscript[W, net]))/m)  ⟹   (Subscript[v, f])=v(30(QuantityUnit))=(Sqrt[(Subscript[v, i]^2)+((2 (Subscript[W, net]))/m)])=5.00(((QuantityUnit))/((QuantityUnit)))