Last Name: First Name: Student ID: TA: Problem 1: (25 points) A block rests on a horizontal frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an angle ! above the horizontal. After the block is pulled through a distance of D = 1.50 m, its speed is 2.60 m/s and 50.0 J of work has been done on it.

1. What is the angle !? 2. What is the mass of the block? 3. If the floor is not frictionless and the coefficient of kinetic friction is " = 0.4, what has to be the magnitude of the tension if the speed of the block is 2.60 m/s after 1.50 m (for this question the angle ! is 30° and the mass of the block 10.0 kg) ?

T = 45.0 N

! v = 0 v = 2.60 m/s

D = 1.50 m Solution: 1. We start with a coordinate system and a free-body diagram: T = 45.0 N v = 2.60 m/s ! y N

v = 0 x

mg D = 1.50 m Three forces are acting on the box: The normal force N: perpendicular to the displacement, the normal force does NO work.

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The weight mg: perpendicular to the displacement, the normal force does NO work. The tension T: the x component of the tension does some work, it is the only source of work in the system. W = T • D = T cos! D ! cos! = W / (T D) = 0.74 ! ! = 42.2°. 2. We use the work kinetic energy theorem. The block starts at rest, the initial kinetic energy is zero. The only work done on the block is from the tension (50 J): ½ mvf

2 = W ! m = 2 W / v2 = 14.8 kg. 3. Now we have to include the work done by the friction. We call F the sliding friction. We start with a free-body diagram: T = 45.0 N

F

v = 2.60 m/s v = 0

! y N

x mg D = 1.50 m Four forces are acting on the box: The normal force N: perpendicular to the displacement, the normal force does NO work. The weight mg: perpendicular to the displacement, the normal force does NO work. The tension T: the x component of the tension does some work. The kinetic friction F: in the opposite direction of the displacement The work done by tension: WT = T • D = T cos! D The work done by friction: WF = F • D = - F D = - " N D = - " (mg – T sin!) D We use the work – kinetic energy theorem: ½ mvf

2 = WT + WF

! ½ mvf2 = T cos! D - " (mg – T sin!) D

! T = (1/2 mvf2 + " mg D) / [D(cos! + " sin!)]

! T = 57.9 N

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Last Name: First Name: Student ID: TA: Problem 2: (25 points) A 1.2 kg block is held at rest against a spring with a force constant k = 730 N/m. The block is NOT attached to the spring. Initially, the spring is compressed a distance D from its relaxed position. When the block is released, it slides across a surface that is frictionless except for a rough patch of width L = 5.0 cm that has a coefficient of kinetic friction " = 0.44. 1. Find D such that the block’s speed after crossing the rough patch is 2.3 m/s. 2. After crossing the rough patch, the 1.2 kg block collides with a 2.4 kg block. The collision is perfectly inelastic. What is the speed of the blocks? 3. What minimum length should the patch be to stop the 1.2 kg block? y Patch length L

x Solution: 1. The only force that does work during the release is the elastic force from the spring, it is a conservative force. The two other forces, normal force and weight, are perpendicular to the displacement, they do not do work on the block. We can use the conservation of mechanical energy, we call v1 the speed after release, the sum of kinetic and potential energy is constant : 0 + ½ kD2 = ½ mv1

2 + 0 For the patch, we use the work kinetic energy theorem, we call v2 the speed of the block after the patch : ½ mv2

2 – ½ mv12 = - " mg L

We have two equations that give : ½ mv2

2 = ½ kD2 - " mg L ! D = [2 (½ mv22 + " mg L) / k]1/2 = 0.097 m

2. The collision is perfectly inelastic, it means that the two blocks coalesce (stick together). We apply the conservation of momentum before and after the collision. Before collision the total momentum is the momentum of the 1.2 kg block, in the x direction, magnitude 1.2 2.3 = 2.76 kg

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m/s. After collision, the total momentum is the momentum of the two blocks stuck together, in the x direction, magnitude : (1.2 + 2.4) v. Conservation of momentum gives : 2.76 = 3.6 v ! v = 7.66 10-1 m/s, in the x direction. 3. We use the condition we found in question 1 with v2 = 0, the work done by friction on the patch equals the elastic energy : ½ mv2

2 = ½ kD2 - " mg L ! ½ kD2 - " mg L = 0 ! L = (1/2 kD2) / (" mg) = 0.66 m

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Multiple Choice Questions (2.5 points per question). 1. A 4.0 kg mass is moving with speed 2.0 m/s. A 1.0 kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?

(a) The 4.0 kg mass. (b) The 1.0 kg mass. (c) Both travel the same distance. (d) Cannot b determined from the information given.

Answer C, the kinetic energy for the two masses is the same ½ mv2 = 8.0 J. If the force applied on both is the same, then by definition of the work kinetic energy theorem, the distance to stop them will be the same. 2. The ratio of the kinetic energy of object A to the kinetic energy of object B is 2:1. Object A is moving with a speed of 6.0 m/s and object B is moving with a speed of 2.0 m/s. If the mass of object A is 4.0 kg then what is the mass of object B?

(a) 2.0 kg (b) 4.0 kg (c) 8.0 kg (d) 18 kg (e) 16 kg

Answer D, using the expression for kinetic energy ½ mv2. 3. A 220 kg object is lifted at a constant speed through a height of 42 m by a crane. If the crane takes a total time of 1.8 seconds to lift the object, what is the average power required? Use g = 10 m/s2.

(a) 25 kW (b) 51 kW (c) 75 kW (d) 100 kW (e) 150 kW

Answer B. The speed is constant, the force provided by the crane equals the weight of the object. The power provided is by definition: P = F • v = mg v = 2200 42 / 1.8 = 51 kW. 4. A roller coaster of mass 80.0 kg is moving with a speed of 20.0 m/s at position A as shown in figure 1. The vertical height at position A above ground level is 200 m. Neglect friction and use g = 10.0 m/s2. What is the speed of the roller coaster at point B?

(a) 66.3 m/s

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(b) 20.0 m/s (c) 46.0 m/s (d) 17.6 m/s (e) 12.5 m/s

Answer A. We use the conservation of mechanical energy, the potential energy from point A to point B is converted into kinetic energy. ½ mvB

2 = ½ mvA2 + mgH, H = 200 m.

160 m

C

200 m

B

A Potential energy = 0 Figure 1 5. Referring to question 4 and figure 1, what is the speed of the roller coaster at point C?

(a) 0 m/s (b) 34.6 m/s (c) 69.2 m/s (d) 20.0 m/s (e) 12.9 m/s

Answer B. We use the conservation of mechanical energy, the potential energy from point A to point C is converted into kinetic energy. ½ mvB

2 = ½ mvA2 + mgH, H = 40 m.

6. An 8.0 kg object moving with an initial velocity of 8.0 m/s on a surface comes to rest due to friction after it travels a horizontal distance of 11 m. What is the coefficient of kinetic friction between the object and the surface? Use g = 10 m/s2.

(a) 0.13 (b) 0.25 (c) 0.29 (d) 0.43 (e) 0.80

Answer C, the normal force equals the weight in magnitude. The sliding friction is the product of the coefficient of sliding friction and the normal force. The work done by sliding friction

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removes the kinetic energy of the object, using the work kinetic energy theorem we have that: ½ mv2 = " mg L ! " = ½ v2 / (gL) = 0.29. 7. An object with a mass of 10 kg is moving along a horizontal surface. At a certain point it has a speed of 28 m/s. If the coefficient of friction between the object and the surface is 0.40, how far will the object go before it comes to a stop? Use g = 10 m/s2.

(a) 98 cm (b) 61 m (c) 98 m (d) 23 m (e) 61 cm

Answer C, the normal force equals the weight in magnitude. The sliding friction is the product of the coefficient of sliding friction and the normal force. The work done by sliding friction removes the kinetic energy of the object, using the work kinetic energy theorem we have that: ½ mv2 = " mg L ! L = ½ v2 / (g ") = 98 m. 8. A force F = (4i + 5j + 1k) Newtons acts on an object that moves from r1 = (5i - 8j + 2k) meters to a new position r2 = (-5i + 4j + 5k) meters. How much work is done on the object? (a) 23 J (b) 87 J (c) -87 J (d) 12 J (e) - 48 J Answer A, the displacement is !r = r2 – r1 = (-10i +12j + 3k), the work done is the dot product of the force and the displacement: W = F • !r = -40 + 60 + 3 = 23 J. 9. A 0.140 kg baseball is dropped from rest from a height of 2.00 m above the ground. What is the magnitude of its momentum just before it hits the ground?

(a) 0.280 kg m/s (b) 0.877 kg m/s (c) 0.439 kg m/s (d) 0.620 kg m/s (e) 1.37 kg m/s

Answer B. The conversion of potential energy to kinetic energy gives the speed of the ball before it hits the ground: mgH = p2 / (2m) ! p = (2m2gH)1/2 = 0.877 kg m/s. 10. A 0.140 kg baseball is dropped from rest from a height of 2.2 m above the ground. It rebounds to a height of 1.6 m. What change in the ball’s momentum occurs when the ball hits the ground?

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(a) 0.117 kg m/s upwards (b) 0.117 kg m/s downwards (c) 1.70 kg m/s upwards (d) 0.350 kg m/s upwards (e) 0.350 kg m/s downwards

Answer C. Momentum is a vector quantity, the initial momentum is oriented downwards, the final momentum is oriented upwards. The change in momentum is the final momentum minus the initial momentum, therefore the change in momentum is oriented upwards and the magnitude is the sum of the initial and final momentum. We use the same reasoning as in question 9. Initial momentum: p = (2m2gH)1/2 = 0.919 kg m/s. Final momentum: p = (2m2gH)1/2 = 0.784 kg m/s. Change in momentum: #p = 1.70 kg m/s upwards.

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Phys 1301.300 - Practice Hour Exam 3 B Date: October 26, 2007

Problem

A spring has a force-distance relationship of F = !10x + 2x3. The equilibrium points are...

Solution

(c) 0, ±"

5Equilibrium points exist where F = 0. This is equivalent to saying 2x3 = 10x. This is true forx = 0 or x2 = 5.

Problem

A 100 kg basketball player hanging on the rim of a basketball hoop deflects the rim by 15 cm.If the force constant of the rim is 7.2 kilonewtons per meter, what is the potential energy storedin the deflection of the rim?

Solution

(c) 81 JIf you consider that the basketball hoop behaves like a typical spring (F = !kx), and you letk = 7200N/m and x = 0.15m, then the standard relation for the energy stored in a spring(PE = 1

2kx2) should hold. Plugging in the values, we get 81 J.

Problem

Which of the following statements is true?

1. The potential energy associated with a conservative force depends only on position.

2. Points at which the derivative of the potential energy function is zero are called stableequilibrium points.

3. Doing work on a system always increases the system’s potential energy.

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Solution

(d) Statements 1. and 2.Statements 1. and 2. are just rewordings of the definitions of conservative forces and equilibriumpoints. You can show that statement 3. is invalid since there can exist situations where the workdone on an object increases its potential (such as putting a book onto a higher shelf).

Problem

Which of the following is a Taylor series about x = 0 that sums to ex?

Solution

(c) 1 + x + x2

2 + x3

6 ...

The definition of a Taylor series is!!

n=0f(n)(a)

n! (x ! a)n, where f (n)(a) is the nth derivative ofthe function you are expanding about the point x = a. For f(x) = ex, this expansion works outto 1 + x1

1! + x2

2! + x3

3! ...

Problem

A force represented by the vector F = 3i + 2j causes a displacement represented by the vectorx = 4i! 3j, where i and j are orthogonal unit vectors lying in a plane. If the work done in thissituation is 6 joules, what is the angle between the force and the displacement?

Solution

(e) 71"

The work done here can be written as W = F ·x. The product of the two vectors can be writtenin two ways: A · B = AxBx + AyBy or A · B = |A||B| cos(!). Using these two relations weget F · x = (3)(4) + (2)(!3) and F · x =

"32 + 22

"42 + 32 cos(!). Equating these gives us a

relationship for the angle: cos(!) = 6/(5"

13).

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