In the figure, two wooden blocks each of 0, 19 kg mass are connected by a string

Q: In the figure, two wooden blocks each of 0, 19 kg mass are connected by a string

let a be the acceleration of the masses for B y = voy*t+ (1/2)*a*t^2 0.8 = 0 + (1/2)*a*2.3^2 a = 0.302 m/s^2 for mass A along horizontal Fnet= mA*a T - uk*mA*g = mA*a .......(1) for mass B mB*g - T = mB*a ........(2) adding 1 +2 mB*g - uk*mA*g = a*(mA+mB) 0.19*9.8 - (uk*0.19*9.8) = 0.302*(0.19+0.19) uk = 0.938

ID: 1562622 • Letter: I

Question

In the figure, two wooden blocks each of 0, 19 kg mass are connected by a string that passes over a very light frictionless pulley. One block slides on a horizontal table, while the other hangs suspended by the string, as shown in the figure. At time t = 0, the suspended block is 0, 8 m over the floor, and the blocks arc released from rest. After 2, 3 s, the suspended block reaches the floor. What is the coefficient of kinetic friction between the table and the sliding block? Answer in two decimal places.

Explanation / Answer

let a be the acceleration of the masses

for B

y = voy*t+ (1/2)*a*t^2

0.8 = 0 + (1/2)*a*2.3^2

a = 0.302 m/s^2

for mass A

along horizontal

Fnet= mA*a

T - uk*mA*g = mA*a .......(1)

for mass B

mB*g - T = mB*a ........(2)

adding 1 +2

mB*g - uk*mA*g = a*(mA+mB)

0.19*9.8 - (uk*0.19*9.8) = 0.302*(0.19+0.19)

uk = 0.938 <<<<<<---------answer

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In the figure, two wooden blocks each of 0, 19 kg mass are connected by a string

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In the figure, two wooden blocks each of 0, 19 kg mass are connected by a string

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