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In the figure, the coefficient of friction between the floor and body \(B\) is \(0.1.\) The coefficient of friction between bodies \(B\) and \(A\) is \(0.2.\) A force \(F\) is applied as shown on \(B.\) The mass of \(A\) is \(m/2\) and of \(B\) is \(m.\)

             

(a) The bodies will move together if \(F = 0.25\text{mg}\)
(b) The \(A\) will slip with \(B\) if \(F = 0.5\text{mg}\)
(c) The bodies will move together if \(F = 0.5\text{mg}\)
(d) The bodies will be at rest if \(F = 0.1\text{mg}\)
(e) The maximum value of \(F\) for which the two bodies will move together is \(0.45\text{mg}\)

Which of the following statement(s) is/are true?
1. (a), (b), (d), (e)
2. (a), (c), (d), (e)
3. (b), (c), (d)
4. (a), (b), (c)

Hint: Apply Newton's laws of motion.
 
Step 1: Find the combined acceleration.
Consider the adjaænt diagram. The frictional force on \((f_1)\) and frictional force on \((f_2)\) will be as shown.
Let \(A\) and \(B\) are moving together the common acceleration is given by;
\(a_\text{common}=\frac{F-f_1}{m_A+m_B}\)
\(a_\text{common}=\frac{2(F-f_1)}{3m}~~~...(1)\)
 
Step 2: Find the pseudo force on the block \(A.\)
The pseudo force on the block \(A\) is given by;
\(F_{pseudo}=(m_A)\times a_\text{common}\)
\(F_{pseudo}=(m_A)\times \frac{2(F-f_1)}{3m}\)
 
                    
The force \((F)\) will be maximum when;
The pseudo force on \(A\) = The frictional force on \(A\)
\(\Rightarrow (m_A)\times \frac{2(F-f_1)}{3m} = \left(μm\right)_{A} g \)
\( = 0 .2 \times \frac{m}{2} \times g = 0 .1 mg \)
\(\Rightarrow F_{max} = 0 .3 mg + t_{1}\)
\( = 0 .3 mg + \left(\right. 0 .1 \left.\right) \frac{3}{2} mg = 0 .45 mg\)

Step 3: Analyse each option using Newton's second law of motion.
Hence, the maximum force unto which bodies will move together is \(F_{max}   =   0 . 45   mg\)
(a) Hence, for \(F = 0 .25 mg < F  _{max}\) bodies will move together. 
(b) For \( F = 0 .5 mg > F_{ max }, \) body \(A\) will slip with respect to \(B.\)
(c)  For \( F = 0 .5 mg > F _{max}  ,\) bodies slip. 
\(f_{1max}=\mu m_B g \Rightarrow 0.1\times \frac{3m}{2} \times g=0.15mg\)
\(f_{2max}=\mu m_A g \Rightarrow 0.2\times \frac{m}{2} \times g=0.1mg\)
Hence, the minimum force required for the movement of the system \( (A + B)\)
\(f_{min}=f_{1max}+f_{2max}\)
\(\Rightarrow f_{min}=0.15mg+0.1mg=0.25mg\)
(d) Given, force \( F = 0 .1 mg < F_{min }\)
Hence, the bodies will be at rest. 
(e) The maximum force for combined movement  \(F_{max} = 0 .45 mg \)
Hence, option (1) is the correct answer.

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