3/4 points | Previous Answers Tipler6 7.P054 My Notes Ask Your Teacher A pendulu
ID: 1875420 • Letter: 3
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3/4 points | Previous Answers Tipler6 7.P054 My Notes Ask Your Teacher A pendulum consists of a 2.0-kg bob attached to a light 2.1-m-long string. While hanging at rest with the string vertical, the bob is struck a sharp horizontal blow, giving it a horizontal velocity of 4.8 m/s. At the instant the string makes an angle of 22 with the vertical, calculate the following: (a) the speed 4.476 m/s (b) the gravitational potential energy (relative to its value is at the lowest point) 2.997 (c) the tension in the string? 16.97 (d) What is the angle of the string with the vertical when the bob reaches its greatest height? 63.88 Draw a sketch of the situation. Let the system be Earth and the pendulum bob. Apply conservation of mechanical energy to the system. When the bob reaches the given angular position its energy will be partially kinetic and partially potential. When it reaches its maximum height, its energy will be entirely potential. Applying Newton's second aw will allow you to express the tension in the string as a function of the bob's speed and its angular position. eBookExplanation / Answer
(a)
intial energy Ei = (1/2)*m*v1^2
final energy Ef = (1/2)*m*v2^2 + m*g*L*(1-costheta)
from energy conservation
Ef = Ei
(1/2)*m*v2^2 + m*g*L*(1-costheta) = (1/2)*m*v1^2
(1/2)*2*v2^2 + (2*9.8*2.1*(1-cos22)) = (1/2)*2*4.8^2
v2 = 4.476 m/s
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(b)
PE = m*g*L*(1-costheta) = (2*9.8*2.1*(1-cos22)) = 2.997 J
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(c)
Tension T = m*g*costheta + mv2^2/L
T = (2*9.8*cos22) + 2*4.476^2/2.1
T = 37.2 N
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(d)
final energy at greatest height Ef = m*g*L*(1-costheta)
Ef = Ei
m*g*L*(1-costheta) = (1/2)*m*v1^2
2*9.8*2.1*(1-costheta) = (1/2)*2*4.8^2
theta = 63.88
DONE please check the answer. any doubts feel free to ask
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