Multiple choice Problem 1

A 50.0-N box is sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the box is sliding to the right at 1.75 m/s and that it stops in 2.25 s with uniform acceleration. The force that friction exerts on this box is closest to:

a) 3.97 N b) 490 N c) 50.0 N d) 8.93 N e) 38.9N Mass of box ism = 50N ÷ 9.8m / s2 = 5.1kg fk v0x = 1.75 m/s ax

Problem 2 A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. She pushes with a force that is much greater than is required to hold the eraser. The force of friction exerted by the board on the eraser increases if she: WARNING: The correct answer may surprise you. Think about the amount of static friction in each case. A) pushes with slightly greater force B) pushes with slightly less force C) pushes so her force is slightly downward but has the same magnitude D) stops pushing E) pushes so her force is slightly downward but has the same magnitude The three diagrams below shows the possible scenario. In all cases the normal force is FN = Mgcos! (! = 0 for A and B). The maximum static friction is fs

max = FNµs . We will assume that in all cases, the upward (+ y, as indicated below) vertical force cancels the downward forces so that the eraser do not fall. A and B C E

fs +y fs fs Static Friction +x Fcosθ !F θ

!F

Applied Fsinθ Fsinθ θ Force

!F Fcosθ

Mg Mg Mg If the eraser is to remain on the wall (not fall) the y-component of the net force must be zero: Fy

net = 0 . The static friction fs < fsmax is calculated for all scenarios below.

ANSWER: Clearly E!

m

Final velocity (x) vx = 0 vx = v0x ! axt,vx = 0ax = v0x / t = 1.75m / s ÷ 2.25s = 0.778m / s

2nd law taking left as positive Fxnet = fk = max = 5.1kg( ) .778m / s2( ) = 3.97N

answer a)

A and B y-component Fynet = fs ! Mg = 0fs = Mg

C, y-component +y (shown above) Fynet = fs ! Mg + F sin" = 0fs = Mg ! F sin"

E, y-component +y (shown above) Fynet = fs ! Mg ! F sin" = 0fs = Mg + F sin!

Problem 3

A series of weights connected by massless cords are given an upward acceleration of 4 m/s2 by a pull P as shown below. A, B and C are the tensions in the connecting cords. The smallest of the three tensions A,B, and C is closest to: a. 483 N pull P b. 621 N c. 196 N d. 276 N 5.00 kg e. 80.0 N

C 10.0 kg B 15.0 kg A

20.0kg

It is obvious that Tension A, TA, between 15.0kg and 20.0 kg mass, is the smallest tension. To solve see free body diagram on 20.0 kg (m = 20.0kg) below: ANSWER: D TA a = 4.0 m/s2 Use 2nd law Fy

net = TA ! mg = ma

TA = m(g + a) = 20kg 13.8m / s2( )

mg TA = 276N Kinematics with Calculus Problem 4 Kinematics with Calculus The position of a particle moving in an xy plane is given by

!r = 2t4 ! 3( ) i + t5 ! 2t( ) j ,

with !r in meters and t in seconds. A) find the average velocity and acceleration for the

time interval between t = 1s and t = 3s. B) find the velocity and acceleration at t = 1s in unit-vector notation. C) What is the angle between the positive direction of the +x axis and a line tangent (i.e.

!v ) to the particle's path at t = 1 s? Give your answer in the range of (-180o; 180o).

A) !r t( ) = xi + yj = 2t4 ! 3( ) i + t5 ! 2t( ) j

at t = 1s, !r 1s( ) = 2 1s( )4 ! 3( ) i + 1s( )5 ! 2 1s( )( ) j = !1mi !1mj

at t = 3s, !r 3s( ) = 2 3s( )4 ! 3( ) i + 3s( )5 ! 2 3s( )( ) j = 159mi + 237mj

!vavg =!x!ti +

!y!t

j =x 3s( ) " x 1s( )3s "1s

i +y 3s( ) " y 1s( )3s "1s

j = 80 msi +119 m

sj

!v t( ) = vxi + vy j =dxdti +

dydt

j =d 2t 4 ! 3( )

dti +

d t 5 ! 2t( )dt

j = 8t 3( ) i + 5t 4 ! 2( ) j

at t = 1s,

!v 1s( ) = 8 1s( )3( ) i + 5 1s( )4 ! 2( ) j = 8 ms i + 3msj

at t = 3s,

!v 1s( ) = 8 3s( )3( ) i + 5 3s( )4 ! 2( ) j = 216 ms i + 403msj

!aavg =!vx!t

i +!vy!t

j =vx 3s( ) " vx 1s( )

3s "1si +

vy 3s( ) " vy 1s( )3s "1s

j = 104 ms2i + 200 m

s2j

b)

!v 1s( ) = 8 1s( )3( ) i + 5 1s( )4 ! 2( ) j = 8 ms i + 3msj

!a t( ) = axi + ay j =dvxdt

i +dvydt

j =d 8t 3( )dt

i +d 5t 4 ! 2( )

dtj = 24t 2( ) i + 20t 3( ) j

at t = 1s,

!a 3s( ) = 24 1s( )2 i + 20 1s( )3 j = 24 ms2i + 20 m

s2j

C)

!v 1s( ) = 8 1s( )3( ) i + 5 1s( )4 ! 2( ) j = 8 ms i + 3msj

vx > 0 and vy > 0 ,1st quadrant ( !900 < " < 0! ).

! = tan"1 3m

8m= 20.5! , ! = 20.5!

Work by Graphical Integration Problem 5

The figure gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set

by as = 5.0ms2

.

A) Calculate the work done after the particle move from xi = 0m to xf = 1.0m. Repeat for xi = 0.5 m to xf = 1.0m. Repeat for the interval xi = 0 to xf = 5.0m. B) If at x = 1m, the particle is at rest (at v = 0), calculate the speed of the particle when it is at xf = 5.0m.

W = Fdxxi

x f! = m adxxi

x f! = m area under the graph[ ]

For xi = 0 to xf = 1.0m:

adx0

1m! = =

12base( ) ! height( ) = 1

21m( ) as( ) = 1

21m( ) 5 m

s2"#$

%&'= 2.5 m

2

s2

HINT: For part a) Use F = ma , and equation on work in the back. For part b) use the equation on power in the back.

W = m adxxi

x f! = 3kg " 2.5 m2

s2= 7.5J

For xi = 0.5m to xf = 1.0m: Here I will enlarge the figure for clarity as as/2

adx0.5m

1m! = = minus

0m 1m 0.5m 1 m 0.5m 1.0m

adx0.5m

1m! =

12

base( ) height( )big "12

base( ) height( )small

= 12

1m( ) 5 ms2

#$%

&'("

12

0.5m( ) 2.5 ms2

#$%

&'(= 1.875 m

2

s2

W = m adx0.5m

1.0m! = 3kg "1.875 m

2

s2= 5.625J

For xi = 0 to xf = 5.0m:

adx0

5m! = + +

=121m( ) 5 m

s2!"#

$%&+ 3m( ) 5 m

s2!"#

$%&+121m( ) 5 m

s2!"#

$%&= 20 m

2

s2

W = m adxxi

x f! = 3kg " 20 m2

s2= 60J

B) For xi = 1m to xf = 5.0m:

adx0

5m! = +

= 3m( ) 5 ms2

!"#

$%&+121m( ) 5 m

s2!"#

$%&= 17.5 m

2

s2

W = m adxxi

x f! = 3kg "17.5 m2

s2= 52.5J

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

2 "12mvi

2 , vi = 0 at x = 1m. The

speed at x = 5m is W = 52.5J = 12mvf

2 ! vf =2 " 52.5J3kg

= 5.9 ms

Angular Momentum Problem 6 In diagram below a 2kg rock is at point P traveling horizontally with a speed of 12 m/s. At this instant what is the magnitude and direction of the angular momentum? If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of the angular momentum?

Angular Momentum

!L = !r ! !p = m!r ! !v valid for point particle w.r.t. point O

!r , r = 8m

36.9° 143.1°

!v For the magnitude

L =

!L = mvr sin143.1" = 2kg !12m / s ! 8m ! .6

L= 115.2kg-m2/s. Torque due to gravity on particle w.r.t. point O.

!! = !r "

!Fg ,Fg = mg = 2kg " 9.8m / s

2 = 19.6N !r , r = 8m

53.1° 36.9°

!Fg

Since the rate of change of angular momentum !! = d

!L / dt has opposite direction (+z)

compared to the direction of the current angular momentum !L = !r ! !p = m!r ! !v (-z), the

angular momentum is decreasing. Can you see the similarity with our much earlier discussion on linear kinetics? FINAL COMMENT AND ADVICE: Angular momentum

!L and Torque

!! depend on

the origin (O). For example, !L = 0 and

!! = 0may be zero for origin O, but nonzero

!L ! 0 and

!! " 0 in another origin O/. Study angular momentum problems and static

equilibrium problems.

Direction perpendicular to x-y plane ! indicates +z out of the page ! indicates –z into page Also +x right +y up

Using the right hand rule on

!L = !r ! !p = m!r ! !v it is easy to see that the direction of

!L is –z or into the page

Using the right hand rule on !! = !r "

!Fg it is easy to see that the

direction of !! is +z or out the page.

! = rFg sin53.1! = 8m "19.6N " .8 = 125.4N im

This can be expressed in a different unit ! = 125.4kg •m2 / s2 Using second law for rotation in terms of angular momentum

!! =

d!Ldt

. Hence the net torque is the rate of change of angular momentum