following statements Next 3 questions (Question S to Question T) are based on th
ID: 1790149 • Letter: F
Question
following statements Next 3 questions (Question S to Question T) are based on the t is thrown vertically down from a height of 20 m from the e earth surface the earth surface as your reference object ie ce, and assume 5 natancal energy of the oect at its point of release is 400 784J 834 e more information is required to answer this question 6. Kinetic energy at a height of S m is 638J b. 588 J c 834 J d. 196 e. None of the above 7. At what height of the object its kinetic energy will be equal to its potential energy? a. 20 m b. 10.6 m C. 10 m d. 7.1 m e. Never For a block of mass m to slide without friction up the rise of height h shown, it must have a minimum initial kinetic energy of: 8. a. 2mgh b. gh c. mgh d. gh/2 e. mghi2 9. An object is constrained by a cord to move in a circular path of radius 0.5m on a horizontal frictionless surface. The cord will break if its tension exceeds 16N. The maximum kinetic energy of the object can have is: a. 64 J b. 32J c. 16J d. 8J e 4J PHY 211 Quiz 3 Version C Page 2 of 3Explanation / Answer
Q-1)
Work done = F*DeltaS
= (4i + 2j +5k) *[ (2i + 9j +7k) - (-3i + 4j + 2k) ]
= (4i + 2j +5k) *[ (5i + 5j +5k) ]
= 20+10+25
= 55 J (option d)
Q-2)
KE = 1/2 mv^2
A) KE = 1/2 2m 4v^2 = 4 mv^2
b) KE = 1/2 2m (4 v)^2 = 16 mv^2
c) KE = 1/2 2m (v)^2 = m v^2
d) KE = 1/2 m (3V)^2 = 4.5 m v^2
so answer is (B)
Q-3)
Work = ?KE+?PE since the cart finished at the same height as it began ?h = 0 thus ?PE = 0
W = KEf - KEi
W = 0.5mvf2 - 0.5mvi2
W = 0.5*m*(vf2 - vi2)
W = 0.5*6*(102 - 62)
W = 0.5*6*64
W = 160J (option a)
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