Download Understanding Work & Kinetic Energy: 1D Analysis & Work-Energy Theorem and more Lecture notes Kinematics in PDF only on Docsity!
7. Work and Kinetic Energy
A) Overview
This unit introduces two important new concepts: kinetic energy and work. These concepts are defined in terms of the fundamental concepts from dynamics ( force and mass ) and kinematics ( displacement and velocity ). We will find that integrating Newton’s second law through a displacement will result in an equation that links these two concepts of kinetic energy and work. This equation (called the work-energy theorem or sometimes, the center of mass equation), allows us to easily answer many questions that would be very difficult using Newton’s second law directly
B) Work and Kinetic Energy in One Dimension
We begin our introduction of work and energy by considering the simple one- dimensional situation shown in Figure 7.1. A force is applied to an object, causing it to accelerate. We say that the force acting over time causes the change in velocity. We can
quantify this statement by integrating the force over time to obtain the relationship between this integral of the force over time and the change in velocity.
dt
dv Fnet = m
2
1
2
1
v
v
t
t
Fnet dt m dv
We could also describe this situation by saying that the force acting through a distance caused the change in velocity. How do we quantify this description? Well, consider the motion at time t : in the next instant of time dt , the velocity will change by an amount dv which is equal to the acceleration times dt. dv = adt In this same time dt , the position will change by an amount which is equal to the velocity dx = vdt times dt. We apply Newton’s second law, to replace the acceleration by the net force divided by the mass and then combine the equations to eliminate dt and obtain the equation:
v
dx m
F
dv = net
Figure 7. Aconstant force Fnet is applied to a an object over a time interval (from t 1 to t 2 ) that results in a change in its velocity and in its displacement.
If we now integrate this equation, we obtain the relationship between the integral of the net force over the displacement and the change in the square of the velocity.
∫ = ∫
2
1
2
1
x
x
net
v
v
m vdv F dx
2
1
1 2 x x
m v Fnet dx
We define the integral of the force over the displacement to be the work done by the force
≡ (^) ∫
2
1
x
x
W Fdx
and the quantity ½ the mass times the velocity squared to be the kinetic energy of the
2 2
K ≡ mv
particle. Thus we see that in one dimension, the work done on the object by the net force is equal to the change in that object’s kinetic energy. Work and kinetic energy both are measured in Joules, where 1 Joule is defined to be 1 N-m.
We will do a simple example in the next section that illustrates the use of these concepts.
C) Example
Figure 7.2 shows a box of mass 6 kg that is initially at rest. A horizontal force of magnitude 24 N is now applied and the box begins to move. We would like to determine the speed of the box when it is at a distance of 8 m from its initial position.
How do we go about making this calculation? We could first use Newton’s second law to determine the acceleration, and then use this acceleration in one of our kinematics equations to determine the time it takes to travel 8 m, and then use another kinematics equation to determine the speed at this time.
There is, however, an easier way. Namely, we can simply equate the work done by the net force (which is just the applied force in this case) to the change in the kinetic energy of the box. W =∆ K
Figure 7. A constant force o 24 N is applied to a box of mass 6 kg that is initially at rest. What is the speed of the box when it reaches a distance of 8m from its starting point?
Figure 7.4 shows two vectors A and B , defined in terms of their x and y components. If
we now take the dot product of these two vectors, we see that the only terms that survive are the products of the same components
A • B = Ax Bx + AyB y
In the next section, we will use the dot product to define the work done by a force
E) Work-Kinetic Energy Theorem
In an earlier section we determined that in one dimension, the integral of the net force over the displacement of an object is proportional to the change in the square of its velocity We will now make use of the dot product to generalize this result to more than one dimension.
We start by defining the work done by a force on an object as the integral of the dot product of the force and the displacement. We can see from this definition that only
≡ (^) ∫ • ℓ
W F d
the part of the force that is parallel to the displacement contributes to the calculation of the work done by the force. For example if you pull a box across a horizontal frictionless floor, both the weight and the normal force are perpendicular to the displacement so that neither of these forces does any work on the box. Only the tension force has a component in the direction of the displacement, so only the tension force does any work on the box.
Figure 7. The representation of vector A and B in terms of their components ( Ax, Ay ) and ( Bx,By ) using the dot product and the unit vectors along the ( x , y ) axes.
( i ˆ,ˆ j )
Figure 7.5 shows a box being pulled up a frictionless ramp so that we have displacements in two dimensions, x and y. We can now basically repeat the derivation of the one
dimensional case for each component. That is, we start from the equations for the change in velocity and displacement during a time dt.
dv a dt
dv a dt
y y
x x
dy v dt
dx v dt
y
x
Once again we apply Newton’s second law to replace the acceleration by the net force divided by the mass and combine the equations to eliminate dt and obtain two equations, one for each dimension. mv (^) x dvx = Fnetx dx mv (^) y dvy = Fnetydy
We now integrate each equation and add them to obtain our final result:
m [ ∫ vx dvx +∫ vydvy ] =∫( Fnetxdx + Fnetydy )
The left hand side of the equation is equal to the change in kinetic energy of the box. The right hand side of the equation is the work done by the net force, since the dot product can be expanded as the sum of the products of the components. ∆ K = W net
We now see that defining the work in terms of the dot product of the force and the displacement allows us to generalize our one-dimensional result. The change in the kinetic energy of an object is equal to the work done on that object by the net force. We call this statement the Work Kinetic Energy Theorem. In the next few sections we will explore the work done by several forces.
Figure 7. A free-body diagram for a box of mass m being pulled up a a ramp by a taut rope.
The net force here is certainly not constant! Therefore, we cannot use our kinematic equations for constant acceleration. Indeed, finding the position and velocity
of the skidding ball at any time is difficult. However, we can use the work-kinetic energy theorem to easily find the final velocity.
Only two forces act on the skidding ball: the normal force provided by surface, and the weight of the ball. Now, the ball always moves parallel to the plane. The normal force is always perpendicular to the plane. Therefore, work done by the normal force, given by the dot product of the normal force and the displacement is zero! Consequently, the work done by the net force here is just equal to the work done by the weight: W =− mg ∆ y
Applying the work-kinetic energy theorem, we see that the final speed of the skidding ball is exactly the same as the final speed for the dropped ball! 2 2 2
1 − mg ∆ y = mv f
v (^) 2 f = 2 g ∆ y
We will close this unit with a discussion of two more examples of the work done by conservative forces.
G) Work Done by a Variable Force: Spring
We have calculated the work done by the weight, a constant force, when the orientation of the path relative to the force was changing. We will now calculate the work done by a spring in one dimension. In this case, the orientation of the path relative to the force is simple, but the magnitude of the force changes as we move.
We define the origin or the coordinate system to correspond to the e relaxed length of the spring as shown in Figure 7.8. The force exerted by the spring on the object attached to its end as a function of position is proportional to the extension or
Figure 7, Two balls are dropped at the same tiem from the same height. Ball 1 falls freely while Ball 2 skids down a frictionless curved surface..
compression of the spring. As we move the object between two positions x 1 and x 2 , the force on the object clearly changes. Breaking the movement into tiny steps we see that
the work done by the spring along each step will depend on the position : To find the total work done we need to integrate this expression between x 1 and x 2.
1 2 21 (^2212 )
2
1
W k xdx k x x
x
x
→ =− ∫ =−^ −
Figure 7.9 shows a plot of the force as a function of displacement as the spring is moved from x 1 to x 2. The area under the curve represents the work done by the force.
Note that the formula for the work done by a spring that we just derived depends only on the endpoints of the motion, x 1 and x 2. In fact, if we first stretch the spring from its equilibrium position, we see that the spring force points in the opposite direction to the displacement and therefore does negative work. If we then compress the spring back to its equilibrium position, the spring force has the same direction, but the displacement is now in the opposite direction so that the spring force does positive work. In fact the magnitudes of these negative and positive works are equal, so that we see that when the spring is returned to its original position, the net work done is zero, the signature of a conservative force.
Figure 7. A spring with spring constant k exerts a restoring force on an object of mass m. The coordinate system is defined such that x = 0 corresponds to the relaxed length of the spring. To calculate the work done by the spring as the object moves between two positions, x 1 and x 2 , we must integrate the varying restoring force from x 1 to x 2 ..
Figure 7. A plot of the force exered by the spring in Fig 7.8 as a function of the extension of the spring. The work done by the spring as the object is moved from x 1 to x 2 is given by the shaded area under the curve.
The total work is therefore just the sum of the work done by all of the radial steps. To make this sum, we do the integral and obtain our result: that the work done by the
(^221) 1 2
1 r^ r
dr GM m r
GM m W (^) E
r
r
E
gravitational force as an object is moved from r 1 to r 2 is proportional to 1/ r 2 – 1/ r 1. Once again we see that the work depends only on the endpoints and not on the specific path, demonstrating that the gravitational force is indeed a conservative force.
Figure 7. To evaluate the work done by gravity, we approximate the curved path as a series of infinitesimal elements each of which are in turn broken down into two elements, one radial and one tangential. The work done along each tangential element is zero since the force is radial. Consequently, the work done by gravity only depends on r 1 and r 2 , the radial distances of the endpoints of the path.
