Block 1 (m1 = 10 kg) is resting on top of block 2 (m2 = 4 kg) on the floor of an
ID: 1771551 • Letter: B
Question
Block 1 (m1 = 10 kg) is resting on top of block 2 (m2 = 4 kg) on the floor of an elevator. The elevator is moving downward with a decreasing speed and the magnitude of the acceleration is 5 m/s2. On a sheet of paper, draw the free body diagrams for block 1 and block 2 using the two-subscript notation from class. After completing the free body diagram, enter below each force and its x & y-components. Remember that the x-component is the "i" component and the y- component is the "j" component. NET force on Block 1 Fmet1 = 0 i + 50 j N NET force on Block 2 Fmet2 = 0 i + 20 j N You are correct. Computer's answer now shown above. Your receipt no. is 163-5518 revious Tries FORCES on BLOCK 1 Weight force on block 1 by Earth Normal force on block 1 by block 2 N12 = 0 i + 150 j N Incorrect. Computer's answer now shown above Previous Tries 212
Explanation / Answer
NET force on Block1
Fnet1 = m1*a = 10*5 j = 50 j
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NET force on Block2
Fnet2 = m2*a = 4*5 j = 20 j
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FORCES on BLOCK1
Weight force on block1
W1E = 0i - m1*g j = 0i - 10*10 j = 0i - 100 j
normal force on block1 by block2
N12 = 0i + 100 j
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FORCES on BLOCK 2
Weight force on block 2 by Earth
W2E = 0i - 40 j
Normal force on block 2 by 1
N21 = -100 j
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Normal force on block 2 by surface
N2s = (m2+m1)*g j = 140 j