Chapter 4

Motion With a Changing Velocity

P2.27: Find the magnitude and direction of the vector with the following components:

(a) x = -5.0 cm, y = +8.0 cm

(b) Fx = +120 N, Fy = -60.0 N

(c) vx = -13.7 m/s, vy = -8.8 m/s

(d) ax = 2.3 m/s2, ay = 6.5 cm/s2

P2.56: A 3.0 kg block is at rest on a horizontal floor. If you push horizontally on the block with a force of 12.0 N, it just starts to move.

(a) What is the coefficient of static friction?

(b) A 7.0-kg block is stacked on top of the 3.0-kg block. What is the magnitude F of the force acting horizontally on the 3.0-kg block as before, that is required to make the two blocks start to move?

P3.47: A 2010-kg elevator moves with an upward acceleration of 1.50 m/s2. What is the tension that supports the elevator?

P3.48: A 2010-kg elevator moves with a downward acceleration of 1.50 m/s2. What is the tension that supports the elevator?

Kinematic Equations for Const. Acceleration

Fnet = ma.

If Fnet is const, a will also be const.

Uniformly accelerated motion: a = const.

A car moves at a constant acceleration of magnitude 5 m/s2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at

(i) t = 2s? (ii) t = 4s? (iii) t = 10s?

A car moves at a constant acceleration of magnitude 5.7 m/s2. At time t = 0, the magnitude of its velocity is 18.3 m/s. What is the magnitude of its velocity at

t = 2.2s?

Kinematic Equations for Const. Acceleration

Consider an object on which a net force Fnet acts on it. Thus it moves with an acceleration.As the object moves, its velocity changes.

Fnet

a

Fnet

aTime = 0

Initial position = x0

Initial velocity = v0

Time = t

Final position = x

Final velocity = v

Fnet

Kinematic Equations for Const. Acceleration

• aave = ainst

• Let us use initial time, t1 = 0.

•Final time, t2 = t, hence t = t – 0 = t

• Position: initial, x1 = x0, final, x2 = x

• Velocity, initial v1 = v0, final, v2 = v

Fnet = ma.

If Fnet is const, a will also be const.

Uniformly accelerated motion:

a = const.

Kinematic Equations for Constant Acceleration

v = v0 + atx = x0 +vot + ½ at2

v2 = v02 + 2a(x-x0)

Average velocity vav = (v0 + v)/2

Uniformly accelerated motion: a = constant.

Time: Initial = 0, final = tPositions: Initial = x0, final = x Velocity: Initial = v0, final = v

Example Problem 4.14 A train traveling at a constant speed of

22 m/s, comes to an incline with a constant slope. While going up the incline the train slows down with a constant acceleration of magnitude 1.4 m/s2.

(a)Draw a graph of vx versus t.

(b)What is the speed of the train after 8.0s on the incline?

(c) How far has the train traveled up the incline after 8.0 s?

A car moving south slows down with at a constant acceleration of 3.0 m/s2. At t = 0, its velocity is 26 m/s. What is its velocity at t = 3 s?

A. B. C. D. E.

0% 0% 0%0%0%

A. 35 m/s south

B. 17 m/s south

C. 23 m/s south

D. 29 m/s south

E. 17 m/s north

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A car initially traveling at 18.6 m/s begins to slow down with a uniform acceleration of 3.00 m/s2. How long will it take to come to a stop?

A. B. C. D. E.

0% 0% 0%0%0%

A. 55.8 s

B. 15.6 s

C. 6.20 s

D. 221.6 s

E. None of these

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Free Fall

• Free fall: Only force of gravity acting on an object making it fall.

• Effect of air resistance is assumed negligible.

• Force of gravity acting on an object near the surface of the earth is F = W = mg.

• Acceleration of any object in free fall:

a = g = 9.8 m/s2 down (ay = -9.8 m/s2).

Free Fall

+y

+x

ay = -9.8 m/s2

ax = 0

Free Fall contd…

1. a = g, regardless of mass of object.

2. a = g, regardless of initial velocity

+y

+x

ay = -9.8 m/s2, ax = 0

v0 = 0 v0 = -15 m/s v0 = +15 m/s

3. Free Fall: Motion is symmetric.

+y

+x

ay = -9.8 m/s2, ax = 0

v0 = +5 m/s

At the maximum height:

• vy = 0

• Speed at equal heights will be equal.

• Equal time going up and down.

Example: Problem 4.32

A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.5 m above the ground when launched.

(a) How high above the ground does the stone rise?

(b) How much time elapses before the

stone hits the ground?

APPARENT WEIGHT

A physics student whose mass is 40 kg stands inside an elevator on a scale that reads his weight in Newtons.

Scale Reading = Normal force the scale

exerts on the student.Scale Reading = N

= mg

= 40 x 9.8 N

= 392 N

N

mg

1. Elevator at rest. What will be the scale reading?

Fnet = N – W = may

At rest means ay = 0.

Hence N = W, ie apparent weight = true weight = 40 x 9.8 = 392 N

N

W = mg

2. Elevator accelerating upwards with ay = 2.0 m/s2. What will be the scale reading?

2. Elevator accelerating upwards with ay = 2.0 m/s2. What will be the scale reading?

Fnet = N – W = may

ay = + 2.0 m/s2 (positive because acceleration is upwards) . Hence,

N –W = N – mg = may.

N = mg + may = m(g+ay) = 40(9.8 + 2.0) = 472 N

ie, apparent weight is greater than the true weight.

N

W = mg

a

3. Elevator accelerating downwards with ay = 2.0 m/s2. What will be the scale reading?

3. Elevator accelerating downwards with ay = 2.0 m/s2. What will be the scale reading?

Fnet = N – W = may

ay = - 2.0 m/s2 (negative because acceleration is downwards) .

Hence N –W = N – mg = -may.

N = mg - may = m(g - ay) = 40(9.8 - 2.0) = 312 N

ie, apparent weight is less than the true weight.

N

W = mg

a

A 112.0-kg person stands on a scale inside an elevator moving downward with an acceleration of 1.80 m/s2. What will be the scale reading?

A. B. C. D. E.

12%7%

0%0%

81%A. 1299 N

B. 1,098 N

C. 896 N

D. 112 N

E. 0 N

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A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise?

A. B. C. D. E.

5% 3%8%

73%

13%

A. 9.8 m

B. 3.00 m

C. 1.15 m

D. 26.1 m

E. 19.6 m

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WEIGHTLESSNESS

If the elevator was going down with an acceleration ay = g = -9.8m/s2, then

N = m(g-g) = 0

ie, apparent weight = 0

This is “weightlessness” or “zero gravity”

Apparent weight of an object in free fall is zero while its true weight remains unchanged.

Equilibrium

Newton’s 2nd Law: Fnet = ma

For an object in equilibrium: Fnet = 0

Static (v = 0) 0r dynamic (v = constant) eqlbm.

2-dimensions, separate the x and y components and treat the problem as two 1-dim problems.

Fx = max

Fy = may

For equilibrium, Fx = max = 0 and Fy = may = 0

2-Dimensions

• X and Y are INDEPENDENT!

• Break 2-D problem into two 1-D problems.

y

x

EquilibriumDetermine the tension in the 6 m rope if it sags 0.12 m in the center when a gymnast with weight 250 N is standing on it.

x direction: Fx = max = 0

-TL cos + TR cos = 0

TL = TR

3 m.12 my direction: Fy = may = 0

TL sin + TR sin - W = 0

2 T sinWT = W/(2 sin

y

x

312.0

tan 3.2

TL

TR

WTR

Equilibrium on a Horizontal Plane

Object at rest or moving with const. velocityFx = max = 0 and Fy = may = 0

No motion until F = > fsmax

Fx = F - fs = 0 or F = fs.

Fy = N - W = 0 or N = W

Fx = F - fk = 0 or F = fk

Fy = N - W = 0 or N = W

Ffs

W

N

Ffk

W

NSliding with constant velocityObject at rest

Equilibrium on an inclined PlaneAn object at rest on an inclined planeFx = max = 0 and Fy = may = 0

f sN

W

xy

f s

N

W

xy

Wx

W y

W

W.sin

W.cos

f sN xy

W.cosW.sin

Equilibrium on an inclined PlaneAn object at rest on an inclined planeFx = max = 0 and Fy = may = 0

f sN xy

W.cosW.sin

Fy = may = 0N - Wcos = 0 or N = Wcos

Fx = max = 0

fs - Wsin = 0 or fs = Wsin

If angle is increased, the object will eventually slide

down the plane. Sliding will start beyond angle max

At max: fsmax = W.sinmax. But fsmax = sN = s(Wcosmax)

Therefore, sWcosmax = Wsin max

OR s = (Wsinmax)/ Wcosmax ie, s = tanmax

Equilibrium on an inclined Plane

f sN xy

W.cosW.sin

N = Wcos

fs = Wsin

• If angle is increased, the object will eventually slide down the plane.• Sliding will start beyond angle max • At max: fsmax = W.sinmax.

• But fsmax = sN = s(Wcosmax)• Therefore, sWcosmax = Wsin max

• OR s = (Wsinmax)/ Wcosmax ie, s = tanmax

An object at rest on an inclined plane

f k

x

FN

y

W.sin

W.cos

A mass m being pulled uphill by a force F

If m = 510 kg, s= 0.42, k =

0.33, = 15o:(a)Find minimum force F

needed to start the mass moving up.

(b)If the force in (a) is maintained on the mass, what will its acceleration be?

(c) To move the mass with constant speed, what must the value of F be?

A block is at rest on a flat board. The flat board is gently tilted. At what angle will the block start to slide? Assume the coefficient of static friction (s) between the block and the board is 0.48.

A. B. C. D. E.

2% 0% 2%

95%

0%

A. 0.48o

B. 61.3o

C. 28.7o

D. 25.6o

E. 0.00837o

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Position, Velocity and Acceleration

• Position, Velocity and Acceleration are Vectors!

• x and y directions are INDEPENDENT!

0

0

tt

rrv

f

fav

0

0

tt

vva

f

fav

0

0

tt

xxv

f

fx

0

0

tt

yyv

f

fy

0

0

tt

vva

f

xxfx

0

0

tt

vva

f

yyfy

22yx vvv

22yx aaa

x direction y direction

Velocity in Two Dimensions

5 m/s

3 m/s

A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the change in velocity?

y

x

x-direction

vix = 5 m/s

vfx = 3 m/s cos(25)

vx = 3cos(25)–5 =-2.28m/s

y-direction

viy = 0 m/s

vfy = 3 m/s sin(25)

vy = 3sin(25)=+1.27 m/s 22yx vvv m/s 6.2

Acceleration in Two Dimensions

5 m/s

3 m/s

A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the average acceleration? [Assume force of gravity is very small].

y

xx-direction y-direction

s 0.5m/s28.2xa

2m/s 21.5

2m/s 56.4 s 0.5.27m/s1xa

2m/s 54.2

22yx aaa

A wagon of mass 50 kg is being pulled by a force F of magnitude 100 N applied through the handle at 30o

from the horizontal. Ignoring friction, find the magnitude of

(a) the horizontal component of F.

(b) the horizontal component of acceleration.

(c) the normal force exerted on the wagon.

Projectile Motion

A projectile – An object moving in 2-dimensions near the surface of the earth with only the force of gravity acting on it.

Eg: golf ball, batted base ball, kicked football, soccer ball, bullet, etc.

• Assume no air resistance.

• Assume g = -9.8 m/s2 constant.

• We are not concerned with the process that started the motion!

Free Fall: 1-dimensional motion.

+y

+x

ay = -9.8 m/s2, ax = 0

v0 = +5 m/s

At the maximum height:

• vy = 0

• Speeds at equal heights will be equal.

• Equal time going up/down.

+y

+x

ay = -9.8 m/s2,

ax = 0

v0 = 5 m/s

PROJECTILE: Free Fall motion in 2-dimensions.

v0x

v0y

PROJECTILE: Free Fall motion in 2-dimensions.

ay = -9.8 m/s2

ax = 0

v0x = v0cos

v0y = v0sin

What will happen to the y-component of the velocity?

What will happen to the x-component of the velocity?

Kinematics in Two Dimensions

• x = x0 + v0xt + ½ axt2

• vx = v0x + axt

• vx2 = v0x

2 + 2ax (x - x0)

• y = y0 + v0yt + ½ ayt2

• vy = v0y + ayt

• vy2 = v0y

2 + 2ay (y – y0)

x and y motions are independent!They share a common time t.

Kinematics for Projectile Motion

ax = 0 ay = -g

• x = x0 + v0t

• vx = v0x

• y = y0 + v0yt - 1/2 gt2

• vy = v0y - gt

• vy2 = v0y

2 - 2g yX

Y

PROJECTILE: Free Fall motion in 2-dimensions.

ay = -9.8 m/s2, ax = 0

Once the projectile is in air, the only force acting on it is gravity. Its trajectory (path of motion) is a parabola.

Fnet = ma = -mg

ay = -9.8 m/s2

ax = 0

PROJECTILE: Free Fall motion in 2-dimensions.

ay = -9.8 m/s2 and ax = 0

v0x = v0cos and v0y = v0sin

Two balls A and B of equal mass m. Ball A is released to fall straight down from a height h. Ball B is thrown horizontally. Which ball lands first?

A B

h

ay = -9.8 m/s2

ax = 0

Vo = 0

v0x = 0

v0y = 0

ay = -9.8 m/s2

ax = 0

Vo 0

v0x = Vo

v0y = 0

A

ay = -9.8 m/s2

ax = 0

vo = 0

v0x = 0

v0y = 0

y0 = 0, y = -h

• y = y0 + v0yt + ½ ayt2

• vy = v0y + ayt• vy

2 = v0y2 + 2ay (y – y0)

To find time t, use

y = y0 + v0yt + ½ ayt2

-h = 0 + (0 . t) + ½ (-g)t2

Gives 2h = gt2 and t = (2h/g)

Bay = -9.8 m/s2

ax = 0

Vo 0

v0x = vo

v0y = 0 • y = y0 + v0yt + ½ ayt2

• vy = v0y + ayt• vy

2 = v0y2 + 2ay (y – y0)

To find time t, use

y = y0 + v0yt + ½ ayt2

-h = 0 + (0 . t) + ½ (-g)t2

Gives 2h = gt2 and t = (2h/g)

A flatbed railroad car is moving along a track at constant velocity. A passenger at the center of the car throws a ball straight up. Neglecting air resistance, where will the ball land?1. Forward of the center of the car

2. At the center of the car

3. Backward of the center of the car

Since no air resistance is present, the ball and the train would be moving with the same horizontal velocity, and when the ball is tossed, it is given an additional velocity component in the vertical direction, but the original horizontal velocity component remains unchanged, and lands in the center of the train.

P 4.22A penny is dropped from the

observation deck of the Empire State building (369 m above the ground). With what velocity does it strike the ground? Ignore air resistance.

P 4.36An arrow is shot into the air at an

angle of 60.0o above the horizontal with a speed of 20.0 m/s.

(a) What are the x- and y-components of the velocity of the arrow 3.0 s after it leaves the bowstring?

(b) What are the x- and y- components of the displacement of the arrow during the 3.0-s interval?

A ball is thrown with a speed of 40.0 m/s at 55o above the horizontal. At the maximum height, its speed will be

A. B. C. D. E.

30%

0%

9%

47%

14%

A. 22.9 m/s

B. -9.8 m/s

C. 0 m/s

D. 32.8 m/s

E. 40.0 m/s

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A ball is thrown with a speed of 40.0 m/s at 35o above the horizontal. How long is it in air?

A. B. C. D. E.

8%

23%

25%25%

20%A. 6.69 s

B. 8.16 s

C. 2.34 s

D. 4.08 s

E. 4.68 s

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A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise?

A. B. C. D. E.

8% 8%10%

69%

5%

A. 9.8 m

B. 3.00 m

C. 1.15 m

D. 26.1 m

E. 19.6 m

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Summary of Concepts

• X and Y directions are Independent!– Position, velocity and acceleration are vectors

• F = m a applies in both x and y direction

• Projective Motion– ax = 0 in horizontal direction

– ay = g in vertical direction

50

1. A car initially traveling at a velocity vo begins to slow down with a uniform deceleration of 1.20 m/s2 and comes to a stop in 26.0 seconds. Determine the value of vo.

A. B. C. D. E.

95%

5%0%0%0%

A. 31.2 m/s

B. 21.7 m/s

C. 27.2 m/s

D. 24.8 m/s

E. None of these

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2. A 102.0-kg person stands on a scale inside an elevator moving downward with anacceleration of 1.300 m/s2. What will be his apparent weight?

A. B. C. D. E.

0%7%

0%2%

91%

A. 999.6 N

B. 132.6 N

C. 867.0 N

D. 1132 N

E. 0 N

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3. A ball is thrown with a speed of 27.0 m/s at 35o above the horizontal. At the maximum height, its speed will be

A. B. C. D. E.

2% 0%

16%

82%

0%

A. 27.0 m/s

B. -9.8 m/s

C. 0 m/s

D. 22.1 m/s

E. 15.5 m/s

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4. A ball is thrown with a speed of 32.0 m/s at 50o above the horizontal. How long is it in air?

A. B. C. D. E.

11% 13%

64%

4%7%

A. 4.20 s

B. 6.53 s

C. 3.27 s

D. 2.50 s

E. 5.00 s

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5. A ball is kicked straight up from ground level with initial velocity of 16.6 m/s. How high above the ground will the ball rise?

A. B. C. D. E.

2%

93%

0%2%2%

A. 28.1 m

B. 14.1 m

C. 1.69 m

D. 0.847 m

E. 1.18 m

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6. A block is at rest on a flat board. The flat board is then gently tilted. If the block starts to slide at a tilt angle of 23.8o, what is the coefficient of static friction (s) between the block and the board?

A. B. C. D. E.

0% 0%

89%

9%2%

A. 0.40

B. 0.91

C. 23.8

D. 87.6

E. 0.44

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