Question

A 28.0 kg block is connected to an empty 2.20 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.42 and the coefficient of kinetic friction between the table and the block is 0.35. Sand is gradually added to the bucket until the system just begins to move (Figure 1).aCalculate the mass of sand added to the bucket.bCalculate the acceleration of the system.

Answer 1

When the system is just begin to move then the frictional force acts on the block will be static friction.

For the block the static friction is,

`\begin{gathered} f_s=\mu_sN=\mu_smg=0.42*28*10=117.6N \\ \end{gathered}`

So to move the block the bucket should apply the force of 117.6N to the block,

`\begin{gathered} Force=f_s=mg=(x+2.20)*10 \\ Where,\text{ } \\ x=mass\text{ }of\text{ }added\text{ }sand \\ so, \\ x=(f_s)/(10)-2.2=(117.6)/(10)-2.2=9.56kg \end{gathered}`

So mass of the added sand to the bucket is 9.56 kg.

Now if the block is start moving then the frictional force act on the block will kinetic friction.

So the equation from the above figure is,

So our equation from both the above figure is,

`\begin{gathered} For\text{ }block \\ T-f_k=28a \\ T=28a+f_k\text{ }..........(1) \\ And\text{ }for\text{ }bucket \\ 11.76g-T=11.76a\text{ }.........(2) \\ So,\text{ }by\text{ }putting\text{ }valueof\text{ }T\text{ }from\text{ }equation\text{ }(1)\text{ }in\text{ }equation\text{ }(2)\text{ }we\text{ }get, \\ 11.76g-(28a+f_k)=11.76a \\ 11.76g-28a-f_k=11.76a\text{ }..........(3) \end{gathered}`

`\begin{gathered} f_k=kinetic\text{ }frictional\text{ }force \\ so, \\ f_k=\mu_kN=\mu_kmg=0.35*28*10=98N \end{gathered}`

So,

`\begin{gathered} So,\text{ }from\text{ }equation\text{ }(3),\text{ } \\ a\left(28+11.76\right)=11.76g-f_k=11.76g-98=117.6-98=19.6 \\ So, \\ a=(19.6)/(28+11.76)=(19.6)/(39.76)=0.4929\text{ m/}s^2 \end{gathered}`

So acceleration of the block is 0.4929 m/s^2.