(tp8327) pampalone HW#11 swaminathan (10500) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 3.41 kg ball is attached to a 13.3 N fish line. The ball is released from rest at the horizontal position (0 = 90 ). At what angle (measured from the vertical) will the fish line break? The acceleration of gravity is 9.8 m/s2 . Correct answer: 82.3765. Explanation: In the swing down to the breaking point, energy is conserved: m v2 2 m v2 . 2 g m cos = r m g r cos = At the breaking point, consider the radial forces: Fr = m ar T - m g cos = m v2 r T = 3 m g cos T = arccos 3mg = arccos 13.3 N 3(3.41 kg)(9.8 m/s2 ) 6. 7. 8. 9. mgL 10 kL - 10 g L m g- kL m L L correct 10 L
1
If the ball is allowed to fall, what is its speed when the system is swinging through the vertical and the spring is stretched an L amount L = , where L is the spring's 10 initial, unstretched length? 1. 2. 3. 4. kL - 11 g m kL - 22 g 10 m 10 g - k L2 mg kL m L 10 L 10 L 100
5. Not enough information is given.
L + L
= 82.3765 .
22 g -
kL 10 m
002 (part 1 of 2) 10.0 points A ball of mass m is attached to a spring of negligible mass, and force constant k. The spring is freely pivoted at the opposite end from the ball, and the system is initially at rest and horizontal.
Explanation: Using conservation of energy with E = K + Us + Ug and taking Ug = 0 at the instant the system is vertical we get Ei = Ef
pampalone (tp8327) HW#11 swaminathan (10500) Ki + Usi + Ugi = Kf + Usf + Ugf 1 1 0 + 0 + m g (L + L) = m v 2 + k (L)2 2 2 1 1 L2 11 L = m v2 + k mg 10 2 2 100 2 22 m g L k L - = m v2 10 100 mL kL 22 g - = mv 2 10 10 m 22 g - kL 10 m L =v 10
2
wrong answer, since the mass m is neither moving in a circle nor at a constant speed. 004 (part 1 of 3) 10.0 points A projectile of mass 4.3 kg kg is shot horizontally with an initial speed of 17.5 m/s from a height of 22.7 m above a flat desert surface. The acceleration of gravity is 9.81 m/s2 . a) For the instant before the projectile hits the surface, find the work done on the projectile by gravity. Correct answer: 957.554 J. Explanation: Basic Concept: Wgravity = Ug = m g h Given: m = 4.3 kg h = 22.7 m g = 9.81 m/s2 Solution: Wgravity = (4.3 kg) (9.81 m/s2 ) (22.7 m) = 957.554 J .
003 (part 2 of 2) 10.0 points Having calculated the speed in the first part, now find the acceleration a of the ball at the point where it swings through the vertical position. kL 1. ^ j -g 10 m 2. ^ g - j 3. -^ j v2
11 L 10
correct
kL -g 10 m kL - 2g 4. ^ j 110 m 5. 0 6. ^ 2g - j kL 110 m
7. ^ g j kL 10 m kL 9. ^ j +g 10 m Explanation: From Newton's 2nd Law, 8. ^ j ma = a =^ j Fi = ^ (k L) - ^ (m g) j j kL -g 10 m .
005 (part 2 of 3) 10.0 points b) Find the change in kinetic energy since the projectile was fired. Correct answer: 957.554 J. Explanation: Basic Concept: K = Wgravity Solution: K = 957.554 J .
Note that substituting v 2 from part 1 into v2 v2 = will generate a completely a= r L + L
006 (part 3 of 3) 10.0 points c) Find the final kinetic energy of the projectile.
pampalone (tp8327) HW#11 swaminathan (10500) Correct answer: 1615.99 J. Explanation: Basic Concept: K = Kf - Ki = Given: vi = 17.5 m/s Solution: Kf = K + Ki 1 = K + mvi 2 2 3. 4. 1 1 2 2 m vf - m vi 2 2
3
By how much has the spring stretched (or compressed) in the second situation described? 1. 2. 3m v0 correct 4k 5m v0 2k 7m v0 4k 1m v0 4k
5. None of these 7m v0 2k Explanation: In the first case, the total energy 6. E1 = K1 + K2 + Us 1 1 2 2 = m v0 + m v0 + 0 2 2 2 = m v0 . In the second case, the total energy E2 = K1 + K2 + Us 1 v0 2 1 1 2 = m + m v0 + k 2 2 2 2 2 1 5 2 = m v0 + k 2 . 8 2 Total energy is conserved, so E1 = E2 1 5 2 2 m v0 = m v0 + k 2 8 2 1 3 2 k 2 = m v0 2 8 3m v0 . = 4k The spring could be stretched ( > 0) or compressed ( < 0). 008 10.0 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle with respect to
1 = (957.554 J) + (4.3 kg) (17.5 m/s)2 2 = 1615.99 J . 007 10.0 points
Two boxes of the same mass m are sliding without friction on a horizontal surface, connected by a spring of force constant k. At a certain instant the spring is unstretched and the two boxes are moving in the same direction at the same speed v0 . v0 v0 m L0 At another instant the leading box is moving at the same speed v0 as before, but the trailing box is moving in the same direction at only half its initial speed. v0 v0 2 m L0 + m m
pampalone (tp8327) HW#11 swaminathan (10500) the horizontal. The block starts from rest and the coefficient of kinetic friction is k .
D
4
F
m
k
009 10.0 points A 2.6 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1151 N provided by traction between the wheels and the road. The other is a 937 N resistive force due to various frictional forces. How far must the car travel for its speed to reach 3.3 m/s? Correct answer: 66.1542 m. Explanation: Basic Concepts: 1 2 Wnet = K = Kf - Ki = mvf 2 1 2 since vi = 0 m/s Ki = mvi = 0 J. 2 Fnet = F1 - F2 Wnet = Fnet d cos = Fnet d since = 0 cos = 1. Given: m = 2.6 103 kg F1 = 1151 N F2 = 937 N vf = 3.3 m/s Solution: 1 2 (F1 - F2 ) d = mvf 2 2 mvf d= 2(F1 - F2 ) (2600 kg)(3.3 m/s)2 d= 2(1151 N - 937 N) = 66.1542 m
What is the final speed of the block? 1. v = 2. v = correct 3. v = 4. v = 5. v = 6. v = 7. v = 8. v = 2 (F sin - k N ) D m 2 (F cos - m g sin - k N ) D m 2 (F cos + m g sin ) D m 2 (F cos + m g sin - k N ) D m 2 (F cos - m g sin ) D m 2 (F sin + k N ) D m 2 (F cos - m g sin + k N ) D m
2 cos (F - k N ) D m Explanation: The work done by gravity is Wgrav = m g D cos(90 + ) = -m g D sin . The work done by the force F is WF = F D cos . From the work-energy theorem we know that Wnet = K , WF + Wgrav + Wf riction = Thus vf = 2 (F cos - m g sin - k N ) D . m 1 2 m vf . 2
010 (part 1 of 3) 10.0 points A block is pushed against the spring with spring constant 14 kN/m (located on the lefthand side of the track) and compresses the
pampalone (tp8327) HW#11 swaminathan (10500) spring a distance 5 cm from its equilibrium position (as shown in the figure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 . 14 kN/m 5 cm 521 g 2.4 m 1.4 m =0.5 vx
5
1 1 2 m vx = k d2 - m g 2 2 k d2 2 vx = - 2g m Thus vx = k d2 -2g m (14000 N/m) (0.05 m)2 = 0.521 kg
1/2
- 2 (0.5) (9.8 m/s2 ) (1.4 m) = 7.31153 m/s . 011 (part 2 of 3) 10.0 points What is the horizontal distance x the block travels in the air? Correct answer: 5.11701 m.
x What is the speed v of the block when it leaves the track? Correct answer: 7.31153 m/s. Explanation: Let : g = 9.8 m/s2 , m = 0.521 kg , k = 14000 N/m , v = vx , = 1.4 m , d = 0.05 m , and vf = 10.0249 m/s . vx d
Explanation: Choosing the point where the block leaves the track as the origin of the coordinate system, x = vx t 1 y = - g t2 2 since axi = 0 m/s2 and vyi = 0 m/s. At y = h (below the jump off height), 1 h = - g t2 2 t= -2 h . g (8) (9)
k
h
x Applying Conservation of Mechanical Energy, Ui = Uf + Kf + W since vi = 0 m/s. 1 1 Since K = m v 2 , Us = k d2 , and W = 2 2 mg,
Since x = vx t , we have x = vx t = = k d2 - 2g m - -2 h g
2h k d2 -2g m g
pampalone (tp8327) HW#11 swaminathan (10500) = - (14000 N/m) (0.05 m)2 (0.521 kg) Calculate the spring force constant k. Correct answer: 792.876 N/m. Explanation:
6
- 2 (0.5) (9.8 m/s2 ) (1.4 m) 2 (-2.4 m) (9.8 m/s2 ) = 5.11701 m .
1/2
Let :
012 (part 3 of 3) 10.0 points What is the total speed of the block when it hits the ground? Correct answer: 10.0249 m/s. Explanation: Kf = Ui , since vi = 0 m/s and hf = 0 m, so the vertical motion gives 1 2 m vy = m g h . 2 vy = = vf = = -2 g h -2 (9.8 m/s2 ) (-2.4 m) so
2 2 vx + vy
m = 128 g , h = 65.7 cm , and x = 4.72041 cm .
Using conservation of energy, we have 1 2 k x = m g (h + x) 2 from which k= 2 m g (h + x) x2 2 (0.128 kg)(9.8 m/s2 )(0.657 m + 0.0472041 m) = (0.0472041 m)2 = 792.876 N/m .
= 6.85857 m/s ,
(7.31153 m/s)2 + (6.85857 m/s)2
= 10.0249 m/s . 013 10.0 points A(n) 128 g ball is dropped from a height of 65.7 cm above a spring of negligible mass. The ball compresses the spring to a maximum displacement of 4.72041 cm. The acceleration of gravity is 9.8 m/s2 .
014 10.0 points A spring with a spring-constant 3 N/cm is compressed 31 cm and released. The 8 kg mass skids down the frictional incline of height 37 cm and inclined at a 24 angle. The acceleration of gravity is 9.8 m/s2 . The path is frictionless except for a distance of 0.9 m along the incline which has a coefficient of friction of 0.5 . k = 3 N/cm 8 kg 37 cm 31 cm 24 =
0. 0.
9m
vf
h
5
Figure: Not drawn to scale. What is the final velocity vf of the mass?
x
Correct answer: 1.67281 m/s. Explanation:
pampalone (tp8327) HW#11 swaminathan (10500) Multiplying by 2 and dividing by m gives us Let : g = 9.8 m/s2 = , k = 3 N/cm = 300 N/m , x = 31 cm = 0.31 m , = 0.5 , = 0.9 m , h = 0.37 m , m = 8 kg , and = 24 , 2 Kf = v 2 , so m 2 Kf v= m = 2 (11.1931 J) (8 kg)
7
= 1.67281 m/s .
Consider the kinetic energy of the mass. The mass receives its initial kinetic energy from the potential energy of the spring Ki = Uspring 1 = k x2 2 1 = (300 N/m) (0.31 m)2 2 = 14.415 J .
(1)
It gains kinetic energy because of the potential energy lost in moving down the incline Kgained = Ulost = mgh (2) 2 = (8 kg) (9.8 m/s ) (0.37 m) = 29.008 J . and loses kinetic energy by doing work on the frictional surface Klost = Wf r = m g cos = (0.5) (8 kg) (9.8 m/s2 ) (0.9 m) cos(24 ) = 32.2299 J . (3)
Alternate Explanation: The potential energy at the top of the hill will be converted into kinetic energy at the bottom of the hill minus energy lost due to the nonconservative friction force. The potential energy at the top of the hill consists of the gravitational potential energy plus the potential energy contained in the compressed spring. Combining Eqs. 1, 2, 3, and 4, we have
v=
1 2 m g h + k x2 - m g cos m 2
015 (part 1 of 2) 10.0 points A 380 g particle in a semi-spherical bowl of radius 0.4 m is released from rest at point A at the level of the center of the bowl, and the surface of the bowl is rough. The speed of the particle at B is 2.3 m/s. The acceleration of gravity is 9.8 m/s2 .
A R B
Since energy is concerved, the final kinetic energy is Kf = Us + Ul - Wf r = (14.415 J) + (29.008 J) - (32.2299 J) = 11.1931 J . However, the final kinetic energy is Kf = 1 m v2 . 2 (3)
What is its kinetic energy at B? Correct answer: 1.0051 J. Explanation:
pampalone (tp8327) HW#11 swaminathan (10500) The kinetic energy at B is KB = 1 2 m vB 2 1 = (380 g)(0.001 kg/g)(2.3 m/s)2 2 = 1.0051 J .
8
016 (part 2 of 2) 10.0 points What is the magnitude of the energy lost due to friction as the particle moves from A to B? Correct answer: 0.4845 J. Explanation: The loss of energy is equal to the work done by the force of friction E = +U + K = -m g R + K = -(380 g) (9.8 m/s2 ) (0.4 m) + 1.0051 J = -0.4845 J |E| = 0.4845 J .