Work Energy Power

thomas (amt2529) – HW 3.4 – Pozuc – (10001) vx2 = −2 g h1 p vx = −2 g h1 q = −2 (−9.81 m/s2 ) (1.8 m)

This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment is due October 19th (Monday) at 10 pm. Solutions immediately follow. 001 (part 1 of 3) 10.0 points A block starts at rest and slides down a frictionless track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. 579 g

1

= 5.94273 m/s .

002 (part 2 of 3) 10.0 points What horizontal distance does the block travel in the air? Correct answer: 4.15692 m. Explanation:

b b b b b b

2.4 m

4.2 m

v Let : b

b

b

h2 = −2.4 m .

With the point of launch as the origin, b

b

b

x What is the speed of the ball when it leaves the track? The acceleration of gravity is 9.81 m/s2 .

Let :

Thus x = vx t = vx

h1 h2

h

2 (−2.4 m) −9.81 m/s2

003 (part 3 of 3) 10.0 points What is the speed of the block when it hits the ground?

v b

b

b

b

Correct answer: 9.07766 m/s. b

b

4.2 m Choose the point where the block leaves the track as the origin of the coordinate system. While on the ramp,

1 m vx2 = −m g h1 2

2 h2 g s

= 4.15692 m .

m

Kb = Ut

s

= (5.94273 m/s)

g = −9.81 m/s2 , m = 579 g , and h1 = 1.8 m .

b b b b b b

2 h2 . g

t=

Correct answer: 5.94273 m/s. Explanation:

1 2 gt 2 s

h2 =

Explanation: Let :

h = 2.4 m .

Now choose ground level as the origin. Energy conservation gives us Kf = Ui 1 m vf2 = −m g h 2

thomas (amt2529) – HW 3.4 – Pozuc – (10001) p −2 g h q = −2 (−9.81 m/s2 ) (2.4 m)

vf =

= 9.07766 m/s .

2

hence vf = =

p

q

2 g (h − y) 2 (9.8 m/s2 ) (1.79 m − 0.873 m)

= 4.23948 m/s . Alternate Solution: p vy = −2 g h2 q = −2 (−9.81 m/s2 ) (2.4 m)

= 6.86207 m/s , so q vf = vx2 + vy2 q = (5.94273 m/s)2 + (6.86207 m/s)2

005 (part 1 of 2) 10.0 points A girl swings on a playground swing in such a way that at her highest point she is 3.4 m from the ground, while at her lowest point she is 0.6 m from the ground. The acceleration of gravity is 9.8 m/s2 .

= 9.07766 m/s .

004 10.0 points A ball of mass 0.19 kg is dropped from a height 1.79 m above the ground. The acceleration of gravity is 9.8 m/s2 . Neglecting air resistance, determine the speed of the ball when it is at a height 0.873 m above the ground. Correct answer: 4.23948 m/s. Explanation: Since the ball is in free-fall, the only force acting on it is the gravitational force. Therefore, we can use the principle of constancy of mechanical energy. Initially, the ball has potential energy and no kinetic energy. As it falls, its total energy (the sum of kinetic and potential energies) remains constant and equal to its initial potential energy. When the ball is released from rest at a height h, its kinetic energy is Ki = 0 and its potential energy is Ui = mgh, where the y coordinate is measured from ground level. When the ball is at a distance y above the ground, its kinetic 1 energy is Kf = mv 2 and its potential energy 2 is Uf = mgy. Applying the equation Ki + Ui = Kf + Uf , we obtain 0 + mgh =

1 m v2 + m g y , 2

9m 3.4 m 0.6 m What is her maximum speed? Correct answer: 7.4081 m/s. Explanation: Basic Concepts: Conservation of energy 1 K = m v2 2 Ugr = m g h . Let : r = 9 m , htop = 3.4 m , hbot = 0.6 m .

and

Solution: We can solve this by using the principle of conservation of energy. We need to know the kinetic and potential energies at two points in time. The girl will be moving the fastest when her kinetic energy is largest which occurs when her potential energy is smallest. This means that she will be moving fastest at the bottom of the swing. By conservation of energy, Etop = Ebot Ktop + Utop = Kbot + Ubot 1 1 2 2 m vtop + m g htop = m vbot + m g hbot . 2 2

thomas (amt2529) – HW 3.4 – Pozuc – (10001) Since vtop = 0,

3

1 [3 htop + hbot ] 4 1 = [3 (3.4 m) + (0.6 m)] 4 = 2.7 m .

h1/2 =

1 2 m vbot = m g (htop − hbot ) so 2 vmax = vbot q = 2 g [htop − hbot ] q = 2 (9.8 m/s2 )[(3.4 m) − (0.6 m)] = 7.4081 m/s .

30

9.8 m/s2

1. 3

◦

006 (part 2 of 2) 10.0 points Given: In the following choices, htop is the height of the highest point and hbot is the height of the lowest point. At what height above the ground will the girl be moving at a speed half of her maximum speed?

007 (part 1 of 2) 10.0 points A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s2 .

m

Correct answer: 2.7 m. Explanation: Let h1/2 = the height, where v = v1/2 = 1 vmax . 2

v

6.6 kg What is the speed of the ball when it is in circular motion? Correct answer: 1.91774 m/s.

r

Explanation:

v1

/2

h1/2

v htop hbottom

From conservation of energy, K1/2 + U1/2 = Ktop + Utop 1 2 m v1/2 + m g h1/2 = 0 + m g htop 2   1 1 2 m g h1/2 = m g htop − v . 2 4 max

Let : ℓ = 1.3 m , θ = 30◦ , g = 9.8 m/s2 , m = 6.6 kg .

and

Use the free body diagram below.

θ T

From Part 1, 1 2 m vmax = m g [htop − hbot ] . 2 Substitution yields m g h1/2 = m g htop −

1 m g [htop − hbot ] 4

mg The tension on the string can be decomposed into a vertical component which balances the weight of the ball and a horizontal

thomas (amt2529) – HW 3.4 – Pozuc – (10001) component which causes the centripetal acceleration, acentrip that keeps the ball on its horizontal circular path at radius r = ℓ sin θ. If T is the magnitude of the tension in the string, then Tvertical = T cos θ = m g

(1)

4

Equating both expressions for vball , we have p 2 π ℓ sin θ = vball = g ℓ tan θ sin θ Tperiod s 2 π ℓ sin θ g ℓ sin2 θ = Tperiod cos θ

and Thoriz = m acentrip or T sin θ =

2 m vball . ℓ sin θ

Tperiod = 2 π (2)

T =

mg cos θ

(3)

and substituting (3) into (2) gives m g tan θ =

2 m vball . ℓ sin θ

Solving for v yields p v = g ℓ tan θ sin θ q = (9.8 m/s2 ) (1.3 m) tan 30◦ sin 30◦ = 1.91774 m/s .

008 (part 2 of 2) 10.0 points How long does it take Tperiod for the ball to rotate once around the axis? Correct answer: 2.12963 s. Explanation: Basic Concept: d = vt. Solution: Because the tangential speed of the ball around the circle is constant, we have vball =

s Tperiod

.

s is the distance the ball travels in one revolution, which is the perimeter of the circle of radius ℓ sin θ, therefore, we have s = 2 π ℓ sin θ .

s

ℓ cos θ g

(1.3 m) cos 30◦ 9.8 m/s2 = 2.12963 s .

= 2π

Solving (1) for T yields

s

as sin θ cancels.