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MIDTERM #2, PHYS 1211 (INTRODUCTORY PHYSICS), 5 November 2013 NAME: STUDENT ID: This exam book has 7 pages including an equation sheet on the last page PART I: MULTIPLE CHOICE QUESTIONS (question 1 to 6) For each question circle the correct answer (a,b,c,d or e).

1. (2.5 points) Three books (X, Y, and Z) rest on a table. The weight of each book is indicated. The force of book Z on book Y is:

A) 0 B) 5N C) 9N D) 14N E) 19N X and Y will exert a downward contact force of 4N + 5 N = 9N on book Z. By Newton’s 1st law Z will exert an upward force of 9N on book Y (it is not in contact with book X). ANSWER: C

2. (2.5 points) Two blocks (A and B) are in contact on a horizontal frictionless surface. A 36-N constant force is applied to A as shown. The magnitude of the force of A on B is:

A) 1.5N B) 6.0N C) 29N D) 30N E) 36N Since there is no friction, Newton’s second law gives 36N = mA + mB( )a , giving an

acceleration of a = 1.5 ms2

. Now apply the second law to box B, FAB = mBa = 30N .

ANSWER: D 3. (2.5 points) A block is first placed on its long side and then on its short side on the same

inclined plane. The block slides down the plane on its short side but remains at rest on its long side. A possible explanation is: (Select the one correct answer)

A) the force of gravity is more nearly down the plane in the second case B) the short side is smoother C) the frictional force is less because the contact area is less

Note that the weight of the blocks is in Newton = N

FBA

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D) the center of gravity is higher in the second case E) the normal force is less in the second case

ANSWER: B Answer C is wrong since a smaller area also means a larger normal force per unit area, so that the total normal force ends up being the same. 4. (2.5 points) A 4-kg cart starts up an incline with a speed of 3 m/s and comes to rest 2 m

up the incline. The total work done on the cart is: A) -6J B) -8J C) -12J D) -18J E) need additional information

Use work-energy theorem W net = !K =12mvf

2 "12mvi

2 = 0 " 124kg( ) 3m

s#$%

&'(2

= "18J .

ANSWER: D 5. (2.5 points) In the figure below a ball rolls down a hill. Which of the following

statements is the correct one? (PE gravitational potential energy; KE kinetic energy) A) both KE and PE remain constant. B) its PE decreases and its KE increases. C) both its PE and KE increase. D) both its PE and KE decrease. E) its PE increases and KE decreases ANSWER: B

6. (2.5 points) A 0.5-kg block slides along a horizontal frictionless surface at 2 m/s. It is brought to rest by compressing a very long spring of spring constant 800 N/m. The maximum spring compression is:

A) 0.6 cm B) 3 cm C) 5 cm D) 7 cm E) 8 cm

Since there’s no friction use conservation of mechanical energy initial (KE only)12mv0

2 =12kx2

(Spring PE only): x =.5kg

800N / m2 ms= 0.05m = 5cm . ANSWER: C

PART II: FULL ANSWER QUESTIONS (question 7 to 10) Do all four questions on the provided space. Show all works.

7. (10 points) The diagram below shows a 4.0 kg box moving down a 30° incline with a speed v0 = 6.0 m/s, which is acted on by an external force ofF = 20.0N . The coefficient of kinetic friction between the surfaces is µk = 0.1 , and the coefficient of static friction isµs = 0.3

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a) Draw a free-body diagram of all the forces acting on the box..

fk note direction of friction opposes velocity n F = 20 N Wx (2 point) W = mg 30°

b) Use Newton’s law to find the acceleration (magnitude and direction) of the box.

y-axis

n = mgcos30! = 4.0kg( ) 9.8m /s2( )cos30! = 33.9N (1 point)

fk = nµk = 33.9N( ) 0.1( ) = 3.39N (1 point)

x-axis +x is up and a is up. F + fk ! mgsin 40! = ma (1.5 point)

F + fk !mgsin30! = 20N + 3.39N ! 4.0kg( ) 9.8m /s2( )sin30! = 3.79N = ma

(1.5 point)

a = 3.79N4.00kg

= 0.947m /s2. The acceleration is 0.947 m/s2 up the incline. (1 point)

c) After 2.00s, how far down (or up) the incline does the box move?

x = v0t + 12at 2 = !6.0m /s( ) 2.0s( ) + 1

20.947m /s2( ) 2.0s( )2 = !10.1m . The box moves 10.1 m down

the incline. (2 point)

8. (10 points) In the diagram below block A and B have mass of 4.00 kg and 5.00 kg, respectively, and the coefficient of kinetic friction between block B and the table is µk = 0.25. Assume that Block B is moving right and accelerating to the right.

+y +x

v0 = 6.0 m/s

Wy

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8 (continued) a) Draw a free-body diagram of block A. The diagram should include all forces acting on Block A as well as the direction of its acceleration. Use Newton’s second law to derive an equation that includes the tension in the rope, T, and the acceleration, a.

Using Newton’s Law T (1 point) T ! mAg = !mAa (1 point) Note the sign: only T is positive (up)

a T ! 4.0kg " 9.8m / s2 = ! 4kg( )aT ! 39.2N = ! 4kg( )a

Equation A (1 point) mAg

b) Draw a free-body diagram of block B. The diagram should include all forces acting on Block B as well as the direction of its acceleration. Use Newton’s second law to derive another equation of the tension in the rope, T, and the acceleration, a.

a

(1 point) FN y-com Fynet = FN ! mBg = 0

FN = 5kg " 9.8m / s2 = 49N(1 point)

fk T Friction (kinetic) fk = FNµk = 49N ! 0.25 = 12.25N mBg (1 point)

x-com Fxnet = T ! fk = mBa

T !12.25N = 5kg( )a Equation B (1 point)

c) Part a and b give two equations with two unknowns, T and a. Solve it to find T and a.

Subtract equation B (part b) by equation A (part a) T !12.25N( ) ! T ! 39.2N( ) = 5kg( )a ! ! 4kg( )a( ) (1 point)

9kg( )a = 26.95N ! a = 2.99 ms2

(1 point)

Substitute into Equation A, T = ! 4kg( ) 2.99 ms2

"#$

%&'+ 39.2N = 27.22N (1 point)

Or into equation B, T = 5kg( ) 2.99 ms2

!"#

$%&+12.25N = 27.22N

9. (10 points) The diagram below (on the next page) shows a crate, held at rest,

compressing a spring 1.00 m from its equilibrium position. The mass of the crate is 15.0 kg and the elastic constant of the spring is 988 N/m. The coefficient of kinetic friction between the crate and the floor is µk = 0.20. Initially the crate is located a distance 3.0 m from a 40° incline. There is no friction between the crate and the surface of the incline.

A

B

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A) After crate is released, it moves toward the incline. Use conservation of energy (or work-energy theorem) to find the speed of the crate after it has moved 3.0 m to reach the bottom of the incline.

Use the conservation of energy (with non-conservative force) theorem between position 0 (initial) and 1 (final): K0 +Uel ,0 +Wfriction = K1 +Uel ,1 (1 points)

Uel ,0 +Wfriction =12mv1

2, where we’ve used v0 = 0 and Uel ,1 = 0 (1 points)

It is easy to see Uel ,0 =12kx2 =

12988N / m( ) 1m( )2 = 494J .. (1 point)

Using ( )( )( ) Nsmkgmgnf kkk 4.29/8.90.152.0 2 ==== µµ . And hence Wfriction = ! fk 3.0m( ) = ! 29.4N( ) 3.0m( ) = !88.2N . (1 point) This gives

Uel ,0 +Wfriction =12mv1

2→ v1 =

2 Uel ,0 +Wfriction( )m

=2 494J ! 88.2N( )

15.0kg= 7.36m / s (2 point)

B) Use conservation of energy or the work-energy theorem to find the maximum distance it moves up the incline before sliding down.

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Use the conservation of total mechanical energy between position 1 and 2 K1 +Ugrav,1 = K2 +Ugrav,2 (1 points) 12mv1

2 =Ugrav,2 , where we’ve used v2 = 0 and set Ugrav,1 = 0 . (1 points)

Hence, we have Ugrav,2 = mgh = mgL sin 40! . (1 points)

This gives

12mv1

2 =Ugrav,2 = mgL sin 40!→

L =v12

2gsin 40!=

7.36m / s( )2

2 9.8m / s2( )sin 40! = 4.3m (1 point)

10. (10 points) A block of mass 2.0 kg is sliding up an incline with a speed of 6.0 m/s. It

slides 2.0m up the incline, stops, then slides back down to its starting point. When it reaches its initial starting point its speed is now 5.17 m/s. There is friction between the block and the surface of the incline.

(a) Calculate the change in kinetic energy ΔK during the stated interval.

!K =12mv2

2 "12mv1

2 =122.0kg( ) 5.17 m

s#$%

&'(2"122.0kg( ) 6 m

s#$%

&'(2= "9.27 (3 points)

(b) Calculate the work done by gravity on the block during the stated interval. Here you don’t need to do any calculations, but you must justify your answer. ZERO. Gravity is a conservative force, so now work is done since final and initial positions are same. To see more clearly, let Wx be the x-component of gravity on block, which, of course, remains the same throughout the motion. The work done as the block slides up is - Wx (2.0m) < 0 since gravity opposes upward slide. For the slide down it is Wx (2.0m) > 0 since gravity aids sliding down. The total work by gravity is Wg = - Wx (2.0m) + Wx (2.0m) = 0 (3 points) (c) Using the result of part a and b, and the work-energy theorem, calculate the work done by friction on the block during the interval Hence the only work done is that by friction Wf, and using the work-energy theorem gives

Wf = !K = "9.27J (2 points) (d) Noting that the work done by friction Wf = !"Eth is negative the increase in thermal energy, calculate the friction force on the block. Wf = !9.27J = !"Eth = fk 4m( )# fk = 2.3N (2 point) Note: 2m up plus 2m down

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USEFUL EQUATIONS !A = Ax

"i + Ay

"j + Az

"k , unit vector notation.

Scalar Product: !Ai!B = AxBx + AyBy + AzBz = ABcos!

Solution of quadratic equation, ax2 + bx + c = 0! x ="b ± b2 " 4ac

2a

Kinematics x = x0 + v0xt +12axt

2 , vx = v0x + axt , vx2 = v0x

2 + 2ax x ! x0( )

2/8.9 smg = , Fg = mg Centripetal acceleration arad =v2

r

Newton’s Laws 0=!= FF net!!

(Object in equilibrium) amF net !!

= (Nonzero net force) Friction fs ! µsFN , fk = µkFN . Work-Energy Theorem K = (1 / 2)mv2 , W =

!F •!d = F cos!( )d = F||d (Straight-Line Motion, Constant Force)

Scalar product form: !F = Fx

"i + Fy

"j + Fz

"k , !d = dx

"i + dy

"j + dz

"k , W =

!F •!d = Fxdx + Fydy + Fzdz

W net = !K =12mvf

2 "12mvi

2 (valid if netW is the net or total work done on the object)

Hooke’s Law Fx = !kx . Work and Energy W =!F •!d = F cos!( )d = F||d ;

W net = !K = (1 / 2)mvf2 " (1 / 2)mvi

2 (valid if netW is the net or total work done on the

object);W grav = !mg yf ! yi( ) (gravitational work), W el = ! (1 / 2)kx f2 ! (1 / 2)kxi

2( ) (elastic work) Conservation of Mechanical Energy (only conservative forces are present) Emech =U + K W net = !"U = ! U2 !U1( ) = "K = K2 ! K1 ,U1 + K1 =U2 + K2 ,Ugrav = mgy ,Uel = (1 / 2)kx

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Non-Conservative Forces Wexternal = !Emech + !Eth (Wext work done by external forces, and we set !Eint = 0 ), where !Eth = fkd (thermal energy or negative work done by friction). Using !Emech = !U + !K = U f "Ui( ) + K f " Ki( ) Final energy U f + K f =Ui + Ki +Wext ! fkd initial energy plus external work minus energy loss due to friction

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