Extra Problem of Midterm 2, November 2, 2016 Question 1 The figure below (top of next page) shows trajectories of four artillery shells. Each fired with the same initial speed. Which trajectory remains in the air for the longest time? Circle the right answer. Hint: ask yourself how to throw a ball so that it remains in the air for the longest period.

A) B) C) D) E) All were in the air for the same amount of time

Question 2 In the figure below, a ship approach the dock at 45 cm/s. An important piece of equipment is thrown from the top of a tower to the ship at 15.0 m/s at 60! above the horizontal. The top of the tower is 8.75m above water level. For the equipment to land at the front of the ship, at what distance D from the dock should the ship be when the equipment is thrown?

Diagram below illustrates the problem.

As always begin by finding the x- and y- components of the velocities of the landing equipment (LP):

v0x = v0 cos60! =15m/ s 0.5( ) =7.5m/ s and

v0y = v0 sin60! =15m/ s 3/2( ) =13m/ s

To find the time it takes to reach the bottom of the cliff look at the vertical (y-axis) component: 2

0y0 5.0v gttyy −+= with y0 = 0 and y = -8.75 m. This gives the quadratic

equation −8.75m = 13m / s( )t − 4.9m / s2( )t 2 → 4.9t 2 −13t − 8.75 = 0with the solutions:

t =13± 13( )2 − 4 4.9( ) −8.75( )

2 4.9( ) = 13±18.459.8

→ t = 3.21s,−0.556s . As discussed in class

the positive answer is the physical solution: t = 3.21 s. The Range is found by looking at the horizontal (x-axis) component, where the velocity is constant sm /5.7 v0x = , and Range≡ x = v0xt = 7.5m/ s( ) 3.21s( ) =24.07m . During the t = 3.21 s period it takes the LP to reach the surface of the lake, the boat must travels a distance xB = v0Bt = 0.45m/ s( ) 3.21s( ) =1.44m towards the lighthouse. Using

the above diagram: D= xB +Range =1.44m+24.07m=25.5m .

Question 3 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:

60°

D

v0 = 15 m/s

Range

8.75 m

v0B = 45 cm/s

y0 = 0 x0 = 0 Take the origin to be at the launch point.

xB = D - Range

+x

+y

A) 0 B) 5N C) 9N D) 14N E) 19 Question 4 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 Question 5 A box of textbooks of m = 24.8 kg rests on a loading ramp that makes an angle θ with the horizontal. The coefficient of kinetic friction is µk = 0.25 and the coefficient of static friction is µs = 0.35. A) As the angle θ is increased, find the minimum angle at which the box starts to slip. B) At this angle, find the acceleration once the box has begun to move. C) At this angle, how fast will the box be moving after it has slid a distance 5.0 m along the loading ramp? Use the kinetic equation

A. As the angle θ is increased, find the minimum angle at which the box starts to slip.

y-component Using first law θcosmgn =

x-component If the box doesn’t then the friction force f depicted in the diagram is a static friction force fs ≤ µsn = µsmgcosθ

ax ax

+x

+y n

n f f

mg mg θ

θ

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

When the box begin to slip, the x-component of the weight (mg) equals the maximum static friction force θµθ cossin mgmg s= .

cosθsinθ

=1µs

; tanθ = µs = 0.35 ; θ = 19.3!

B. At this angle, find the acceleration once the box has begun to move.

When !3.19=θ x-component of the weight overcomes the static friction force and the box accelerates down the incline. The friction force opposing the motion will be a kinetic friction force θµµ cosmgnf skk ==

Using the above diagram and Newton’s second law:

xk mamgmg =− θµθ cossin ; θµθ cossin gga kx −= :

ax = 9.8m / s2( )sin19.3! − 0.25 9.8m / s2( )cos19.3! , ax = 0.92m / s2

C. At this angle, how fast will the box be moving after it has slid a distance 5.0 m along the loading ramp? Use the kinetic equation

vx2 = v0x

2 + 2axx = 2axx

vx = 2axx = 2 0.92m / s2( ) 5m( ) = 3m / s

Question 6 In the diagram below block A has a mass of 4.00 kg and block B has mass 12.00 kg. The coefficient of kinetic friction between block B and the table is µk = 0.25. Block B is moving right and accelerating to the right with a = 2.00 m/s2.

(a) Draw a free-body diagram of block A. Hence find the tension in the rope connecting Blocks A and B. (b) Draw a free-body diagram of block B. Calculate the

B

C

a

The rope is massless and there is no friction in the pulley

A

v

friction force on block B. Determine the tension of the rope connecting blocks B and C. (c) Finally, draw a free-body diagram of block C. Determine the mass of block C.

A) Draw a free-body diagram of block A. Hence find the tension in the rope connecting Blocks A and B.

B) Draw a free-body diagram of block B. Calculate the friction force on block B. Determine the tension of the rope connecting blocks B and C.

For x-component use Newton’s 2nd Law: amfTT BkABBC =−− After some rearrangement TBC = TAB + fk + mBa = 47.2N + 29.4N + 12.0kg( ) 2.0m / s2( ) NNTBC 1016.100 == C) Finally, draw a free-body diagram of block C. Determine the mass of block C.

Question 7

y

mAg

TAB

a

Using Newton’s 2nd Law

y

mBg

TBC

a n

TAB ƒk

x Using Newton’s 1st Law for y-axis:

Friction:

y

mCg

TBC

a

Using Newton’s 2nd Law

In the diagram above, block A (100kg) is on a 30° incline is connected by a frictionless pulley to block B (50 kg) on a 53.1° incline. (a) Which way will the system move when released from rest. (b) What is the acceleration of the blocks? (c) What is the tension in the cord?

a) Since there is no friction, to find the acceleration, a, only the components of the forces parallel to the incline (basically the x components). For Box A the x-component of the weight is WAx =WA sin 30

! = mAgsin 30! = 100kg( ) 9.8m / s2( ) 0.5( ) = 490N . For Box B

the x-component of the weight is

30° 53.1°

30° +x

+y

WA

nA

T a

+y +x

T

53.1°

nB

WB

FREE-BODY DIAGRAM of A

FREE-BODY DIAGRAM of B

mA = 100kg mB = 50kg

There is no Friction

a

WBx = −WB sin53.1! = −mBgsin53.1

! = − 50kg( ) 9.8m / s2( ) 0.8( ) = −392N . Since the magnitude of AxW is greater than BxW , box A will accelerate down the incline and box B up the incline, as shown in the diagram.

b) For Box A, using second Newton’s law on the x-component

amTWF AAxnetx =−= →490N −T = 100kg( )a . [1]

For Box B, using second Newton’s law on the x-component

amWTF BBxnetx =−= →T −392N = 50kg( )a . [2]

Adding [1] and [2] 98N = 150kg( )a→a=0.65m/ s2 .

c) Substituting the answer for a into equation [1] we obtain

T = 490N − 100kg( )a = 490N − 100kg( ) 0.65m / s2( ) = 425N .

Alternatively, substitute the value of a into equation [2]

T = 392N + 50kg( )a = 392N − 50kg( ) 0.65m / s2( ) = 425N .

Question 8 Block A of mass mA = 2.0 kg is moving up a 36.9° incline with a speed of v0 = 5.0 m/s. It is attached to block B of mass mB = 4.0 kg, by a massless frictionless rope-pulley system. The coefficient of kinetic friction between Block A and the surface of the incline is µk = 0.2. Block A moves up 0.5 m to the top of incline. Assume block A accelerates up incline and B is accelerating downward as in diagram.

v0 = 5.0 m/s a BLOCK A a BLOCK B 36.9° A) Draw a free-body diagram of block B, then use Newton’s second law to write an equation that includes the tension of the rope, T and the acceleration, a. B) Draw a free body diagram on block A showing all the forces on A and the direction of its acceleration, a. Draw the x-y axes parallel and perpendicular to the incline. C) Use the diagram in part b) to find the normal force (perpendicular to incline) and the force of friction (parallel to incline) on block A. D) Using diagram from part b, determine the x-component (parallel to incline) of the net force, and use the second law to write an equation that include the tension of the rope, T and the acceleration, a. E) The two equations from part a) and d) are two equations with two unknown a and T. Solve them to obtain the tension T and acceleration a.

B

a) Draw a free-body diagram of block B, then use Newton’s second law to write an equation that includes the tension of the rope, T and the acceleration, a. +y Newton’s 2nd law for y (vertical) component Fy

net = T − mBg = −mBa

T − 4.0kg × 9.8m / s2 = − 4kg( )a T (1 point) T − 39.2N = − 4kg( )a , (e1) (1 point) mBg a

b) Draw a free body diagram on block A showing all the forces on A and the

direction of its acceleration, a. Draw the x-y axes parallel and perpendicular to the incline.

+y a +x n T (2 points) fk mAg θ θ = 36.9o

c) Use the diagram in part b) to find the normal force (perpendicular to incline) and

the force of friction (parallel to incline) on block A. y (vertical) component, use diagram of part b)

Fynet = n − mAgcos36.9

! = 0→ n = 2.0kg × 9.8m / s2 × 0.8 = 15.68N Kinetic friction fk = µkn = 0.2 ×15.68N = 3.14N

d) Using diagram from part b, determine the x-component (parallel to incline) of the net force, and use the second law to write an equation that include the tension of the rope, T and the acceleration, a. x (horizontal) component, use diagram of part b)

Fxnet = T − fk − mAgsin 36.9

! = ma Use result of part c) Fx

net = T − 3.14N − 2.0kg × 9.8m / s2 × 0.6 = 2.0kg( )a T −14.9N = 2.0kg( )a , (e2)

e) The two equations from part a) and d) are two equations with two unknown a and T. Solve them to obtain the tension T and acceleration a. Partial answer (a = 4.05m / s2 ). From part a), e1, T − 39.2N = − 4kg( )a→ T = 39.2N − 4kg( )a From part d), e2, T −14.9N = 2.0kg( )a→ T = 14.9N + 2kg( )a Subs. e2 into e1 to eliminate T,39.2N − 4kg( )a = 14.9N + 2kg( )a→ 24.3N = 6.0kg( )a

This give a = 4.05m / s2 Subs. a = 4.05m / s2 into e1 gives T = 39.2N − 4kg( ) 4.05m / s2( ) = 23N

Verify by Subs. a = 4.05m / s2 into e2 gives T = 14.9N + 2kg( ) 4.05m / s2( ) = 23N