An object is gently placed on a long converges belt moving with \[11\,m{s^{ - 1}}\] If the coefficient of friction is \[0.4\], then the block will side in the belt up to a distance of
A.\[10.21m\]
B. \[15.125m\]
C. \[20.3m\]
D. \[25.6m\]
Answer
270k+ views
Hint: In order to solve this problem, first we have to find the retardation produced by the frictional force and then the distance up to which the block will slide by using the equation \[{v^2} = {u^2} + 2as\].
Formula used:
Expression of frictional force,
$f=\mu mg$
Here, $m$ is the mass of the block, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.
The equation of motion is,
\[{v^2} = {u^2} + 2as\]
where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement.
Complete step by step solution:
We are given that \[u = 11\,m{s^{ - 1}}\].
Coefficient of friction \[ = 0.4\]
Now the force of friction between the block and the belt is \[f = \mu mg\].
Now the block will slide on the belt without slipping until its speed (\[v\]) becomes equal to the speed of the belt.
Therefore, \[\mu mg = ma\]
Now this force produces a retardation in the block which is given by,
\[a = -\mu g\]
\[\Rightarrow a = -0 \cdot 4 \times 10\]
\[\Rightarrow a = -4\,m/{s^2}\]
According to the question, the block will slide on the belt without slipping until the speed of the belt.
Now we use the equation \[{v^2} = {u^2} + 2as\], we find the distance of the block
Since \[v = 0\],
So, we have \[{v^2} = {u^2} + 2as\]
By substituting all the values, we get
\[s = -\dfrac{{{u^2}}}{{2a}}\]
\[\Rightarrow s =- \dfrac{{{{\left( {11} \right)}^2}}}{{2 \times(- 4)}}m\]
\[\Rightarrow s = \dfrac{{ 121}}{8}m\]
\[\therefore s = 15.125\,m\]
Thus, the block will slide in the belt up to a distance of \[15.125\,m\].
Hence, option(B) is correct.
Note: The velocity and the distance contribute to the friction evolving between the surfaces. The impact of these two quantities can be resolved. By knowing the velocity and the distance moved by the object, the coefficient of friction can be calculated.
Formula used:
Expression of frictional force,
$f=\mu mg$
Here, $m$ is the mass of the block, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.
The equation of motion is,
\[{v^2} = {u^2} + 2as\]
where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement.
Complete step by step solution:
We are given that \[u = 11\,m{s^{ - 1}}\].
Coefficient of friction \[ = 0.4\]
Now the force of friction between the block and the belt is \[f = \mu mg\].
Now the block will slide on the belt without slipping until its speed (\[v\]) becomes equal to the speed of the belt.
Therefore, \[\mu mg = ma\]
Now this force produces a retardation in the block which is given by,
\[a = -\mu g\]
\[\Rightarrow a = -0 \cdot 4 \times 10\]
\[\Rightarrow a = -4\,m/{s^2}\]
According to the question, the block will slide on the belt without slipping until the speed of the belt.
Now we use the equation \[{v^2} = {u^2} + 2as\], we find the distance of the block
Since \[v = 0\],
So, we have \[{v^2} = {u^2} + 2as\]
By substituting all the values, we get
\[s = -\dfrac{{{u^2}}}{{2a}}\]
\[\Rightarrow s =- \dfrac{{{{\left( {11} \right)}^2}}}{{2 \times(- 4)}}m\]
\[\Rightarrow s = \dfrac{{ 121}}{8}m\]
\[\therefore s = 15.125\,m\]
Thus, the block will slide in the belt up to a distance of \[15.125\,m\].
Hence, option(B) is correct.
Note: The velocity and the distance contribute to the friction evolving between the surfaces. The impact of these two quantities can be resolved. By knowing the velocity and the distance moved by the object, the coefficient of friction can be calculated.
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