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Q. 17 of 75
A 10 kg block is connected to an
empty 1kg bucket by a cord
running over a friction less pulley.
The static friction coefficient is
0.4 and the kinetic friction
between the table and block is
0.3. Sand is gradually added to
the bucket until the system just
begins to move. The mass of
sand added to the bucket and the
acceleration (in m/s2) of the
system respectively (g =
acceleration due to gravity)​

Answer 1

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Answer 2

Answer:

The mass of the sand added to bucket is 3kg and the acceleration of the system is

$0.714ms {}^{ - 2}$

Explanation:

Given that,

The mass of block=10kg

The mass of bucket=1kg

The coefficient of static friction is 0.4.Therefore

$μ _{s} = 0.4$

The coefficient of kinetic friction between table and block is 0.3.Therefore,

$μ _{k} = 0.3$

Let the mass of sand added in bucket be m kg.

The static friction force is

$f _{s}= μ _{s}mg = 0.4 \times 10 \times 10 = 40N$

This force is equal to the weight of bucket filled with sand.Therefore,

$40 = (m + 1)g \\ = > 40 = (m + 1)10 \\ = > m = 3 \: kg$

hence,the mass of sand added is 3kg.

Let the tension in the cord be T.

The kinetic friction force is

$f _{k} = μ _{k}mg \\ = 0.3 \times 10 \times 10 = 30 N$

From free body diagram of bucket,we have

$40-T=4a - - > (1)$

From free body diagram of block we have

$T - 30=10a - - > (2)$

Adding equations 1 and 2 we get

$10 = 14a \\ = > a = \frac{10}{14} = 0.714ms {}^{ - 2}$

Therefore,The mass of sand added to bucket is 3kg and the acceleration of the system is

$0.714ms {}^{ - 2}$

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