a 4 kg block is at rest on the right edge of a large 11kg slab. there is no fric
ID: 1463626 • Letter: A
Question
a 4 kg block is at rest on the right edge of a large 11kg slab. there is no friction below the slab and the coefficient of kinetic friction is .25 between the two masses. the slab is 3m wide and a constant horizontal force is applied to the left on the block. if it takes 2s for the block to reach the other side of the slab, what is the magnitude of the force applied to the block? how far does the bottom mass travel? what is the final velocity of the top and bottom block? (hint: use a non-moving/accelerating reference frame, i.e. static origin on the right side of the table)
Explanation / Answer
uo is constant = 4pi e -7
so
dI/dt = 2 * 6 e -4/( 4*3.14 e-7 * 1000 * 0.4 e-2)
dI/dt = 238.85 A/s
part A : magnitude of force applied F = f = u N = umg
F = 0.25* 4* 9.81
F = 9.8 N
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on bottom Block F = 0.25 * 15* 9.81
F= 36.75 N
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acceration a = F/m = 36.75 /11 = 3.34 m/s^2
use the Kinemtaic eqn S = ut + 0.5 at^2
S = 0.5* 3.34 * 2*2
S = 6.67 m
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final velocity
top block , accleration a = 9.8/4 = 2.45 m/s^2
final speed V = at =2.45 * 2 = 4.9 m/s
Vof bottom block = 3.34 * 2 = 6.68 m/s
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