Week 7 Homework Solutions PDF

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1. Two blocks of mass mA and mB are connected by a very light string passing over a massless and frictionless pulley (see figure). Traveling at constant speed, the block A moves a distance d to the right and block B moves the same distance downward. During this process, how much work is done (a) on block B by gravity, and by the tension in the string? Since the blocks are traveling at constant speed, the net force must be zero. Therefore the tension force acting on B and the weight of B must have equal magnitude (and opposite direction). Since both forces are constant, we can use

𝑊 = 𝐹 ∙ ∆𝑟

For the work done by gravity, the force and the displacement are in the same direction, so that 𝑊𝑔𝑟𝑎𝑣 𝑜𝑛 𝐵 = 𝑚𝐵 𝑔𝑑 For the work done by tension, the force and displacement are in opposite directions, so that 𝑊𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑛 𝐵 = −𝑚𝐵 𝑔𝑑 (b) on block A by gravity, by the tension in the string, by friction, and by the normal force? The gravitational force, and the normal force act on block A in the y-direction. Since there is no motion in this direction, these forces do no work: 𝑊𝑔𝑟𝑎𝑣 𝑜𝑛 𝐴 = 𝑊𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑛 𝐴 = 0 Since the tension is provided by a light string, we can assume that it is constant through. Therefore the tension force acting on A is the same as the tension force acting on B, which we know is equal in magnitude to the weight of B. And since the blocks are traveling at constant speed (in the x-direction) the net force must be zero. Therefore the tension force and the frictional force acting on A must have equal magnitude (and opposite direction). For the work done by tension, the force and the displacement are in the same direction, so that 𝑊tension 𝑜𝑛 𝐴 = 𝑚𝐵𝑔𝑑 For the work done by friction, the force and displacement are in opposite directions, so that 𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑜𝑛 𝐴 = −𝑚𝐵 𝑔𝑑 (c) Find the total work done on each block. For each block we can see that summing the work done by each force gives a total work of zero. We can also see this from the work-energy theorem. Since the speed is constant, the kinetic energy does not change. This means the total work must be zero.

2. A mass m slides down a smooth inclined plane from an initial vertical height h, making an angle α with the horizontal.

(a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height h. From our FBD,

We can decompose the weight into terms parallel and perpendicular to the ramp. The perpendicular component contributes no work, as F · d = 0. The component of the gravitational force acting along the ramp is mg sin α. The length of the ramp it slides down is sinh α . The work is the product of these two: W = mg sin α

h = mgh sin α

(13)

Of course, we could also have gotten this answer by noting that W = F · d = (−mgˆ y) · (−

h x ˆ − hˆ y) = mgh tan α

(14)

We’re not going to wax poetic about this result very much, but this is exactly the work gravity does pulling an object of mass g down a height of −h.

(b) Use the workenergy theorem to prove that the speed of the mass at the bottom of the incline is the same as if it had been dropped from height , independent of the angle of the incline. Explain how this speed can be independent of the slope angle. The work energy theorem tells us that W = ∆KE

(15)

In our system, there is only one source of work, which is gravity. As previously discussed, the work that gravity does is dependent only on the distance the ball travels in y ˆ . It doesn’t care about the slope of the ramp. We can see on the FBD that if we take the ramp to be very long (α very small), then the length over which we do our work increases, but the amount of weight that acts in the direction of motion gets correspondingly smaller.

(c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom. 1 mv2 2 1 g∆h = v2 2 v2 = 2g∆h p v = 2g∆h

mg∆h = W = ∆KE =

For us, this comes out to 17.16 m/s.

(16) (17) (18) (19)

3. A block (mass m) is forced against a horizontal spring of negligible mass and force constant k compressing the spring a distance d. When released, the block slides on a horizontal tabletop with coefficient of kinetic friction µk. Use the work–energy theorem to find (a) how far the block moves from its initial position before coming to rest. First we choose a (one-dimensional) coordinate system:

x 0

-d

We have chosen the origin to be the equilibrium position for the spring. This is generally a good idea, since it allows to characterize the force that the spring exerts on the block as 𝐹𝑠𝑝𝑟𝑖𝑛𝑔 𝑜𝑛 𝑏𝑙𝑜𝑐𝑘 = −𝑘𝑥 Writing the work-energy theorem, we have 𝑊 = ∆𝐾 1

1

𝑊𝑠𝑝𝑟𝑖𝑛𝑔 + 𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 2 𝑚𝑣𝑓2 − 2 𝑚𝑣𝑖2

→

The spring only does work while it is in contact with the block: 0

0 𝑥2 1 𝑊𝑠𝑝𝑟𝑖𝑛𝑔 = ∫ −𝑘𝑥 𝑑𝑥 = −𝑘 [ ] = 𝑘𝑑 2 2 −𝑑 2 −𝑑

Friction does (negative) work on the block throughout its journey: 𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = ∫

𝑥𝑚𝑎𝑥

−𝑑

𝑥𝑚𝑎𝑥 −𝜇𝑘 𝑚𝑔 𝑑𝑥 = −𝜇𝑘 𝑚𝑔[𝑥]−𝑑 = −𝜇𝑘 𝑚𝑔(𝑥𝑚𝑎𝑥 + 𝑑)

Key to solving this problem is to realize that the distance travelled by the block (its maximum displacement), corresponds to the final velocity 𝑣𝑓 being zero. (Note also that the initial velocity 𝑣𝑖 is zero): →

1

2

𝑘𝑑 2 − 𝜇𝑘 𝑚𝑔(𝑥𝑚𝑎𝑥 + 𝑑 ) = 0 − 0

→

(𝑥𝑚𝑎𝑥 + 𝑑 ) =

→

𝑥𝑚𝑎𝑥 =

𝑘𝑑 2 2𝜇𝑘 𝑚𝑔

𝑘𝑑 2 2𝜇𝑘 𝑚𝑔

(this is the total distance travelled)

−𝑑

is the final x-coordinate

(b) the maximum velocity that the block attains. The velocity is maximum when the acceleration is zero (if the acceleration is non-zero then the block is either getting faster (in which case its velocity will reach a maximum sometime in the future), or it is getting slower (which means the velocity was at a maximum sometime in the past). To determine where this occurs we need to analyze the forces.

fk

Fs 𝐹𝑠 − 𝑓𝑘 = 𝑚𝑎

Applying Newton’s II law:

𝑎 = 0 when 𝐹𝑠 = 𝑓𝑘

−𝑘𝑥 = 𝜇𝑘 𝑚𝑔

→

𝑥=

→

−𝜇𝑘 𝑚𝑔 𝑘

This is the position of the block when the velocity is maximum. To find the actual velocity at this point we can use the work energy theorem, calculating the work done on the block as it moves from its initial position to the point at which the velocity it maximum:

𝑊𝑠𝑝𝑟𝑖𝑛𝑔 = ∫

−𝜇𝑘 𝑚𝑔 𝑘

−𝑑

𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = ∫

−𝜇𝑘 𝑚𝑔 𝑘

−𝑑

−𝜇𝑘 𝑚𝑔

𝑘 𝑥2 −𝑘𝑥 𝑑𝑥 = −𝑘 [ ] 2 −𝑑

−𝜇𝑘 𝑚𝑔

−𝜇𝑘 𝑚𝑔 𝑑𝑥 = −𝜇𝑘 𝑚𝑔[𝑥]−𝑑 𝑘 𝑊𝑠𝑝𝑟𝑖𝑛𝑔 + 𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 =

Work-Energy Theorem: →

1

𝑘𝑑 2 − 2

→

𝑣𝑓2 =

→

𝑣𝑓2 =

→

𝑣𝑓 =

=

1

𝜇𝑘2 𝑚2 𝑔2 2𝑘

𝑚𝑘

𝜇𝑘2 𝑚2 𝑔2 𝑘

=

1 𝑚𝑣𝑓2 2

𝜇𝑘2 𝑚2 𝑔2 − 𝜇𝑘 𝑚𝑔𝑑 𝑘 1

− 2 𝑚𝑣𝑖2 1

− 𝜇𝑘 𝑚𝑔𝑑 = 2 𝑚𝑣𝑓2 − 0

(𝑘 2 𝑑 2 − 2𝜇𝑘 𝑚𝑔𝑘𝑑 + 𝜇𝑘2 𝑚2 𝑔2 )

1 (𝑘𝑑 𝑚𝑘 1

+

1 2 𝜇𝑘2 𝑚2 𝑔2 𝑘𝑑 − 2𝑘 2

√𝑚𝑘

− 𝜇𝑘 𝑚𝑔)2

(𝑘𝑑 − 𝜇𝑘 𝑚𝑔)

Now that was a little tricky. Especially recognizing the perfect square in the work-energy theorem. A problem like this might also benefit from a graphical approach:

F Fs = -kx

kd 𝜇𝑘 𝑚𝑔

-d

−𝜇𝑘 𝑚𝑔 𝑘

x

−𝜇𝑘 𝑚𝑔

fk = -μkmg

We have drawn a Force-displacement graph, showing the spring force and the friction force. As the block travels from its starting position (-d) to the point of maximum velocity ( do work, equal to the areas under the curves.

−𝜇𝑘 𝑚𝑔 𝑘

) both these forces

The work done by the spring force (green area) is positive, while the work done by friction (yellow area) is negative. We note that the green area is a triangle on top a rectangle, and that this rectangle has the same area as the yellow rectangle. We can represent the work done as the sum of areas:

Wnet =

+

_

The two rectangles cancel, so that the net Work done is represented by the area of the triangle: 𝑊𝑛𝑒𝑡 =

𝑊𝑛𝑒𝑡 = 𝑊𝑛𝑒𝑡 =

1 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 2

1 𝜇𝑘 𝑚𝑔 + 𝑑) × (𝑘𝑑 − 𝜇𝑘 𝑚𝑔) × (− 𝑘 2

1 1 × (𝑘𝑑 − 𝜇𝑘 𝑚𝑔) × (𝑘𝑑 − 𝜇𝑘 𝑚𝑔) = (𝑘𝑑 − 𝜇𝑘 𝑚𝑔)2 2𝑘 2𝑘

Applying the work-energy theorem (with vi = 0): 1 1 (𝑘𝑑 − 𝜇𝑘 𝑚𝑔)2 = 𝑚𝑣𝑓2 2𝑘 2 →

𝑣𝑓 =

1

√𝑚𝑘

(𝑘𝑑 − 𝜇𝑘 𝑚𝑔)

as before

4. (a) Can the total work done on an object during a displacement be negative? Explain. Yes. If the force and displacement are in opposite directions, then the work done by the force is negative. Negative work will tend to reduce the kinetic energy of the object. For example, think of a kinetic frictional force that is opposing the motion of a sliding object. It acts opposite to the direction of motion so that the work done is negative. This negative work reduced the kinetic energy of the object. If the total work is negative, can its magnitude be larger than the initial kinetic energy of the object? Explain. No. If this were the case the negative work could result in a negative kinetic energy (and there is no such thing!). The maximum amount of negative work that can be done on an object is equal to its kinetic energy. When this work is done the kinetic energy will be reduced to zero (i.e. the object will stop moving). For example, think of an external force applied to a moving object like this:

v Fext This force is in the opposite direction to the motion, and so does negative work. This work will reduce the kinetic energy until it is zero and the object stops. At this moment, the negative work done will be equal to the initial kinetic energy of the object. If we continue to apply the force, it will cause the object to accelerate to the left. The force and the motion will now be in the same direction so that the force is doing positive work.

(b) A car speeds up while the engine delivers constant power. Is the acceleration greater at the beginning of this process or at the end? Explain.

The power delivered by the engine is related to the force by: 𝑃 = 𝐹𝑣 = 𝑚𝑎𝑣 So we can see that the acceleration must decrease as the velocity increases: 𝑎=

𝑃 𝑚𝑣

In a constant power situation, less force is applied at high speed. If this were not the case, and the force remained constant (such as is the case for falling objects), we would find that the power would increase as the same force is being applied over larger and larger distances as the object speeds up.

5. A box of mass m is initially at x0 = 0 sliding along the horizontal +x-axis with velocity v0. It is observed to stop at x1 due to friction.

(a) Express the work done by friction on the box Wf in terms of the frictional force fk and the other given variables. The work done by the frictional force is 𝑓

𝑊𝑓 = ∫ 𝑓𝑘 ∙ 𝑑𝑟 𝑖

The (kinetic) frictional force is constant: 𝑊𝑓 = 𝑓𝑘 ∙ ∆𝑟 and acts in the negative x-direction: 𝑊𝑓 = −𝑓𝑘 𝑖 ∙ ∆𝑟

The displacement ∆𝑟 = 𝑥1 𝑖 is in the positive x-direction:

𝑊𝑓 = −𝑓𝑘 𝑥1 𝑖 ∙ 𝑖 = −𝑓𝑘 𝑥1

(b) Express the friction force fk, in terms of the given variables (no Wf in your answer, please). The work-energy theorem relates the net work done on a system, to its change in kinetic energy: 𝑊𝑛𝑒𝑡 = 𝐾𝑓 − 𝐾𝑖 There are, of course, forces other than friction acting on the box: n

fk x

W

But the weight and the normal force have no component in the direction of motion, so neither force does any work. Therefore the work done by the frictional force is the net work: 𝑊𝑓 = 𝐾𝑓 − 𝐾𝑖 𝑊𝑓 =

1 1 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2 2 2

−𝑓𝑘 𝑥1 =

1 1 𝑚(0) − 𝑚𝑣02 2 2

𝑓𝑘 =

𝑚𝑣20 2𝑥1

(c) Find the time that the box takes to stop, eliminating f, Wf, and m from your answer Apply Newton’s II law to the box (in x-direction) to determine the acceleration of the box: ∑ 𝐹𝑜𝑛 𝑏𝑜𝑥,𝑥 = 𝑚𝑎𝑥 −𝑓𝑘 = 𝑚𝑎𝑥

𝑚𝑣02 = 𝑚𝑎𝑥 2𝑥1 𝑣02 𝑎𝑥 = − 2𝑥1

−

Now use kinematics to find the time that the box takes to stop: ∆𝑥 = 𝑣𝑖 ∆𝑡 +

1 𝑎 ∆𝑡 2 2 𝑥

1 𝑣2 𝑥1 − 0 = 𝑣0 ∆𝑡 + (− 0 ) ∆𝑡2 = 0 2𝑥1 2 ∆𝑡 2 −

4𝑥1 4𝑥12 ∆𝑡 + 2 = 0 𝑣0 𝑣0 ∆𝑡 =

2𝑥1 𝑣0

A person now pushes on the box against the force of friction, sliding it to point x2, where it has speed v1. (d) Find the work done by the person on the box (no Wf in your answer, please).

Start with the work energy theorem: 𝑊𝑛𝑒𝑡 = 𝐾𝑓 − 𝐾𝑖 We must now account for the work done by friction, and the work done by the person: 𝑊𝑓 + 𝑊𝑝𝑒𝑟𝑠𝑜𝑛 = 𝑊𝑝𝑒𝑟𝑠𝑜𝑛 =

1 1 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2 2 2

1 1 𝑚𝑣12 − 𝑚(0) − 𝑊𝑓 2 2

The work done by friction as the box moves from x1 to x2: 𝑊𝑓 = −𝑓𝑘 (𝑥2 − 𝑥1 ) Substituting the expression for fk from part b): 𝑚𝑣02 𝑊𝑓 = − ( ) (𝑥2 − 𝑥1 ) 2𝑥1 Therefore the work done by the person is given by:

𝑊𝑝𝑒𝑟𝑠𝑜𝑛 =

𝑚𝑣02 1 ) (𝑥2 − 𝑥1 ) 𝑚𝑣12 + ( 2𝑥1 2

(𝑥2 − 𝑥 ) 1 1 1 𝑊𝑝𝑒𝑟𝑠𝑜𝑛 = 𝑚𝑣12 + 𝑚𝑣20 𝑥1 2 2...