Questions 26- 28 are based on the following information: A 73S N man is standing
ID: 1865459 • Letter: Q
Question
Questions 26- 28 are based on the following information: A 73S N man is standing on a scale in an elevator 26) What is the reading on the scale in Newtons if the elevator is moving upward with a constant acceleration of 2 meters per second squared? a) 735 N b) 585 N c)75 N d) 150 N e) 885 N 27) What is the reading on the scale in Newtons if the elevator is moving downward with a constant acceleration of 2 meters per second squared ? a) 735 N b) 585 N c)75 N d) 150N e) 885 N 28) What is the reading on the scale in Newtons if the elevator is moving upward with a constant velocity of 5 meters per second? a) 735 N b) 585 N c)75 N d 150N e 885 N Questions 29-33 are based on the following: A mass on an incline shown below starts from rest and slides down a smooth frictionless incline. After it reaches the bottom of the incline, it attains a kinetic energy of 48 joules and then slides on a horizontal frictional surface where the block slides for 6 meters before it come to rest. n- 3.00 kg 0 30.0° 29) What is the height, h, where the block initially starts upon the incline? a) 16m b) 163 m ) 4.89 m d) 10.24 m e) Not enough information given to answer the question 30) What is the velocity of the block when it reaches the bottom of the incline? a) 5.66 m/s b) 32 m/s 16 m/s d) 17.28 m/s e) None of the aboveExplanation / Answer
Answer:
Given that weight of the man is W = 735 N = mg,
from this mass of the man m = 735 N/g = 735 N / 9.8 ms-2 = 75 kg.
(26) If the elevator is moving upward with a constant acceleration a = 2 m/s2 then the total acceleration of the man is a + g = 2 m/s2 + 9.8 m/s2 = 11.8 m/s2
Therefore, the weight of the man is 11.8 m/s2 x 75 kg = 885 N.
(27) If the elevator is moving downward with a constant acceleration a = 2 m/s2 then the total acceleration of the man is g - a = 9.8 m/s2 -2 m/s2 = 7.8 m/s2
Therefore, the weight of the man is 7.8 m/s2 x 75 kg = 585 N.
(28) If the elevator is moving upward with a constant velocity v = 5 m/s then the acceleration is a = 0 m/s2 then the total acceleration of the man is a + g = 0 m/s2 + 9.8 m/s2 = 9.8 m/s2.
Therefore, the weight of the man is 9.8 m/s2 x 75 kg = 735 N.
(29) After the mass reaches the bottom of the incline, it attain a kinetic energy K = 48 J, that means the total potential energy of the mass U is converted completely into its kinetic energy K.
Therefore, U = K or mgh =1/2 mv2
From this, height h = K/mg = 48 J / (3.00 kg) (9.8 m/s2) = 1.63 m.
(30) Velocity of the mass when it reaches the bottom comes from the kinetic energy K = 1/2 mv2 , from this velocity v = (2K/m)1/2 = (2 . 48 J / 3.00 kg)1/2 = 5.66 m/s.
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