A 26.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a

Question

A 26.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley (Fig. 4-57). The coefficient of static friction between the table and the block is0.430 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.

(a) Calculate the mass of sand added to the bucket (Hint: Use the coefficient of static friction.) answer here is 10.18kg

(b) Calculate the acceleration of the system. (Hint: Use the coefficient of kinetic friction.)

_______m/s^2 (downward)

*The answer to (b) isn't 5.9m/s^2

Thanks!

Explanation / Answer

a) 10.18 kg

b) let m1 = 26 kg

m2 = (1 + 10.18)

= 11.18 kg

Let T is the tension in the string.

Let a is the acceleleration of the blocks.

net force acting onm1, Fnet1 = T - mue_k*N

m1*a = T - mue_k*m1*g

T = m1*a + mue_k*m1*g --(1)

net force acting on m2, Fnet2 = m2*g - T

m2*a = m2*g - T

T = m2*g - m2*a --(2)

from equations 1 and 2

m1*a + mue_k*m1*g = m2*g - m2*a

a*(m1+m2) = m2*g - mue_k*m1*g

a = (m2*g - mue_k*m1*g)/(m1+m2)

= (11.18*9.8 - 0.32*26*9.8)/(26+11.18)

= 0.754 m/s^2 <<<<<<<----------------Answer

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