146014
In the arrangement shown in figure \(a_{1}, a_{2}, a_{3}\), and \(a_{4}\) are the accelerations of masses \(m_{1}, m_{2}\), \(m_{3}\), and \(m_{4}\) respectively. Which of the following relation is true for this arrangement?
A Using constraints, \(\Sigma \overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{a}}=0\) \(-4 \mathrm{Ta}_{1}-2 \mathrm{Ta}_{2}-\mathrm{Ta}_{3}-\mathrm{Ta}_{4}=0\) \(4 \mathrm{a}_{1}+2 \mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}=0\)
JEE Main-26.06.2022
Laws of Motion
146015
A cylinder of mass \(12 \mathrm{~kg}\) is sliding on plane with an initial velocity \(20 \mathrm{~ms}^{-1}\). If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of
1 \(40 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(20 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
A Given, \(\mathrm{m}=12 \mathrm{~kg}\) \(\mathrm{u}=20 \mathrm{~ms}^{-1}\) \(\mu=0.5\) We know that Friction force \((\mathrm{f})=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\therefore \mathrm{N}=\mathrm{mg}]\) \(=0.5 \times 12 \times 10\) \(\mathrm{f}=60\) \(\because \quad \mathrm{f}=-\mathrm{ma}\) Acceleration \((a)=-\frac{f}{m}=-\frac{60}{12}=-5 \mathrm{~ms}^{-2}\) From Newton's third law of motion, \(v^{2}=u^{2}+2 a s\) \(0=(20)^{2}+2 \times(-5) \times s\) \(0=400-10 s\) \(s=40 m\) So, cylinder describes distance of \(40 \mathrm{~m}\) before stopping.
AP EAMCET-23.08.2021
Laws of Motion
146016
Two masses \(m\) and \(2 m\) are hang from a frictionless. Weightless ideal pulley as shown below: The upward acceleration of the mass \(m\) is
1 \(\frac{g}{8}\)
2 \(\frac{g}{4}\)
3 \(\frac{g}{3}\)
4 \(\frac{\mathrm{g}}{2}\)
Explanation:
C And \(\quad \begin{aligned} 2 \mathrm{mg}-\mathrm{T}=21 \\ \mathrm{~T}-\mathrm{mg}=\mathrm{ma}\end{aligned}\) By adding equation (i) and (ii) \(2 \mathrm{mg}-\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{a}=\frac{\mathrm{g}}{3}\)
Tripura-2020
Laws of Motion
146017
Two masses \(\mathrm{m}_{1}=5 \mathrm{~kg}\) and \(\mathrm{m}_{2}=4.8 \mathrm{~kg}\) tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? \(\left(\mathrm{g}=9.8 \mathrm{~m} . \mathrm{s}^{-2}\right)\)
1 \(0.2 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(5.0 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(4.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
A Given, \(\mathrm{m}_{1}=5 \mathrm{~kg}\) \(\mathrm{~m}_{2}=4.8 \mathrm{~kg}\) \(\mathrm{~g}=9.8 \mathrm{~ms}^{-3}\) FBD for block \(\mathrm{m}_{1}\) - \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a}\) FBD for block \(\mathrm{m}_{2}-\) \(\mathrm{T}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{2} \mathrm{a}\) On adding equation (i) and equation (ii), we get- \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}+\mathrm{m}_{2} \mathrm{a}\) \(\mathrm{g}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)=\mathrm{a}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{a}=\frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) By putting the value, \(\mathrm{a}=\frac{(5-4.8) \times 9.8}{5+4.8}\) \(\mathrm{a}=0.2 \mathrm{~ms}^{-2}\)
146014
In the arrangement shown in figure \(a_{1}, a_{2}, a_{3}\), and \(a_{4}\) are the accelerations of masses \(m_{1}, m_{2}\), \(m_{3}\), and \(m_{4}\) respectively. Which of the following relation is true for this arrangement?
A Using constraints, \(\Sigma \overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{a}}=0\) \(-4 \mathrm{Ta}_{1}-2 \mathrm{Ta}_{2}-\mathrm{Ta}_{3}-\mathrm{Ta}_{4}=0\) \(4 \mathrm{a}_{1}+2 \mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}=0\)
JEE Main-26.06.2022
Laws of Motion
146015
A cylinder of mass \(12 \mathrm{~kg}\) is sliding on plane with an initial velocity \(20 \mathrm{~ms}^{-1}\). If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of
1 \(40 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(20 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
A Given, \(\mathrm{m}=12 \mathrm{~kg}\) \(\mathrm{u}=20 \mathrm{~ms}^{-1}\) \(\mu=0.5\) We know that Friction force \((\mathrm{f})=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\therefore \mathrm{N}=\mathrm{mg}]\) \(=0.5 \times 12 \times 10\) \(\mathrm{f}=60\) \(\because \quad \mathrm{f}=-\mathrm{ma}\) Acceleration \((a)=-\frac{f}{m}=-\frac{60}{12}=-5 \mathrm{~ms}^{-2}\) From Newton's third law of motion, \(v^{2}=u^{2}+2 a s\) \(0=(20)^{2}+2 \times(-5) \times s\) \(0=400-10 s\) \(s=40 m\) So, cylinder describes distance of \(40 \mathrm{~m}\) before stopping.
AP EAMCET-23.08.2021
Laws of Motion
146016
Two masses \(m\) and \(2 m\) are hang from a frictionless. Weightless ideal pulley as shown below: The upward acceleration of the mass \(m\) is
1 \(\frac{g}{8}\)
2 \(\frac{g}{4}\)
3 \(\frac{g}{3}\)
4 \(\frac{\mathrm{g}}{2}\)
Explanation:
C And \(\quad \begin{aligned} 2 \mathrm{mg}-\mathrm{T}=21 \\ \mathrm{~T}-\mathrm{mg}=\mathrm{ma}\end{aligned}\) By adding equation (i) and (ii) \(2 \mathrm{mg}-\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{a}=\frac{\mathrm{g}}{3}\)
Tripura-2020
Laws of Motion
146017
Two masses \(\mathrm{m}_{1}=5 \mathrm{~kg}\) and \(\mathrm{m}_{2}=4.8 \mathrm{~kg}\) tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? \(\left(\mathrm{g}=9.8 \mathrm{~m} . \mathrm{s}^{-2}\right)\)
1 \(0.2 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(5.0 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(4.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
A Given, \(\mathrm{m}_{1}=5 \mathrm{~kg}\) \(\mathrm{~m}_{2}=4.8 \mathrm{~kg}\) \(\mathrm{~g}=9.8 \mathrm{~ms}^{-3}\) FBD for block \(\mathrm{m}_{1}\) - \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a}\) FBD for block \(\mathrm{m}_{2}-\) \(\mathrm{T}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{2} \mathrm{a}\) On adding equation (i) and equation (ii), we get- \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}+\mathrm{m}_{2} \mathrm{a}\) \(\mathrm{g}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)=\mathrm{a}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{a}=\frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) By putting the value, \(\mathrm{a}=\frac{(5-4.8) \times 9.8}{5+4.8}\) \(\mathrm{a}=0.2 \mathrm{~ms}^{-2}\)
146014
In the arrangement shown in figure \(a_{1}, a_{2}, a_{3}\), and \(a_{4}\) are the accelerations of masses \(m_{1}, m_{2}\), \(m_{3}\), and \(m_{4}\) respectively. Which of the following relation is true for this arrangement?
A Using constraints, \(\Sigma \overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{a}}=0\) \(-4 \mathrm{Ta}_{1}-2 \mathrm{Ta}_{2}-\mathrm{Ta}_{3}-\mathrm{Ta}_{4}=0\) \(4 \mathrm{a}_{1}+2 \mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}=0\)
JEE Main-26.06.2022
Laws of Motion
146015
A cylinder of mass \(12 \mathrm{~kg}\) is sliding on plane with an initial velocity \(20 \mathrm{~ms}^{-1}\). If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of
1 \(40 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(20 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
A Given, \(\mathrm{m}=12 \mathrm{~kg}\) \(\mathrm{u}=20 \mathrm{~ms}^{-1}\) \(\mu=0.5\) We know that Friction force \((\mathrm{f})=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\therefore \mathrm{N}=\mathrm{mg}]\) \(=0.5 \times 12 \times 10\) \(\mathrm{f}=60\) \(\because \quad \mathrm{f}=-\mathrm{ma}\) Acceleration \((a)=-\frac{f}{m}=-\frac{60}{12}=-5 \mathrm{~ms}^{-2}\) From Newton's third law of motion, \(v^{2}=u^{2}+2 a s\) \(0=(20)^{2}+2 \times(-5) \times s\) \(0=400-10 s\) \(s=40 m\) So, cylinder describes distance of \(40 \mathrm{~m}\) before stopping.
AP EAMCET-23.08.2021
Laws of Motion
146016
Two masses \(m\) and \(2 m\) are hang from a frictionless. Weightless ideal pulley as shown below: The upward acceleration of the mass \(m\) is
1 \(\frac{g}{8}\)
2 \(\frac{g}{4}\)
3 \(\frac{g}{3}\)
4 \(\frac{\mathrm{g}}{2}\)
Explanation:
C And \(\quad \begin{aligned} 2 \mathrm{mg}-\mathrm{T}=21 \\ \mathrm{~T}-\mathrm{mg}=\mathrm{ma}\end{aligned}\) By adding equation (i) and (ii) \(2 \mathrm{mg}-\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{a}=\frac{\mathrm{g}}{3}\)
Tripura-2020
Laws of Motion
146017
Two masses \(\mathrm{m}_{1}=5 \mathrm{~kg}\) and \(\mathrm{m}_{2}=4.8 \mathrm{~kg}\) tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? \(\left(\mathrm{g}=9.8 \mathrm{~m} . \mathrm{s}^{-2}\right)\)
1 \(0.2 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(5.0 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(4.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
A Given, \(\mathrm{m}_{1}=5 \mathrm{~kg}\) \(\mathrm{~m}_{2}=4.8 \mathrm{~kg}\) \(\mathrm{~g}=9.8 \mathrm{~ms}^{-3}\) FBD for block \(\mathrm{m}_{1}\) - \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a}\) FBD for block \(\mathrm{m}_{2}-\) \(\mathrm{T}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{2} \mathrm{a}\) On adding equation (i) and equation (ii), we get- \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}+\mathrm{m}_{2} \mathrm{a}\) \(\mathrm{g}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)=\mathrm{a}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{a}=\frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) By putting the value, \(\mathrm{a}=\frac{(5-4.8) \times 9.8}{5+4.8}\) \(\mathrm{a}=0.2 \mathrm{~ms}^{-2}\)
146014
In the arrangement shown in figure \(a_{1}, a_{2}, a_{3}\), and \(a_{4}\) are the accelerations of masses \(m_{1}, m_{2}\), \(m_{3}\), and \(m_{4}\) respectively. Which of the following relation is true for this arrangement?
A Using constraints, \(\Sigma \overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{a}}=0\) \(-4 \mathrm{Ta}_{1}-2 \mathrm{Ta}_{2}-\mathrm{Ta}_{3}-\mathrm{Ta}_{4}=0\) \(4 \mathrm{a}_{1}+2 \mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}=0\)
JEE Main-26.06.2022
Laws of Motion
146015
A cylinder of mass \(12 \mathrm{~kg}\) is sliding on plane with an initial velocity \(20 \mathrm{~ms}^{-1}\). If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of
1 \(40 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(20 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
A Given, \(\mathrm{m}=12 \mathrm{~kg}\) \(\mathrm{u}=20 \mathrm{~ms}^{-1}\) \(\mu=0.5\) We know that Friction force \((\mathrm{f})=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\therefore \mathrm{N}=\mathrm{mg}]\) \(=0.5 \times 12 \times 10\) \(\mathrm{f}=60\) \(\because \quad \mathrm{f}=-\mathrm{ma}\) Acceleration \((a)=-\frac{f}{m}=-\frac{60}{12}=-5 \mathrm{~ms}^{-2}\) From Newton's third law of motion, \(v^{2}=u^{2}+2 a s\) \(0=(20)^{2}+2 \times(-5) \times s\) \(0=400-10 s\) \(s=40 m\) So, cylinder describes distance of \(40 \mathrm{~m}\) before stopping.
AP EAMCET-23.08.2021
Laws of Motion
146016
Two masses \(m\) and \(2 m\) are hang from a frictionless. Weightless ideal pulley as shown below: The upward acceleration of the mass \(m\) is
1 \(\frac{g}{8}\)
2 \(\frac{g}{4}\)
3 \(\frac{g}{3}\)
4 \(\frac{\mathrm{g}}{2}\)
Explanation:
C And \(\quad \begin{aligned} 2 \mathrm{mg}-\mathrm{T}=21 \\ \mathrm{~T}-\mathrm{mg}=\mathrm{ma}\end{aligned}\) By adding equation (i) and (ii) \(2 \mathrm{mg}-\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{mg}=3 \mathrm{ma}\) \(\mathrm{a}=\frac{\mathrm{g}}{3}\)
Tripura-2020
Laws of Motion
146017
Two masses \(\mathrm{m}_{1}=5 \mathrm{~kg}\) and \(\mathrm{m}_{2}=4.8 \mathrm{~kg}\) tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? \(\left(\mathrm{g}=9.8 \mathrm{~m} . \mathrm{s}^{-2}\right)\)
1 \(0.2 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(5.0 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(4.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
A Given, \(\mathrm{m}_{1}=5 \mathrm{~kg}\) \(\mathrm{~m}_{2}=4.8 \mathrm{~kg}\) \(\mathrm{~g}=9.8 \mathrm{~ms}^{-3}\) FBD for block \(\mathrm{m}_{1}\) - \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a}\) FBD for block \(\mathrm{m}_{2}-\) \(\mathrm{T}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{2} \mathrm{a}\) On adding equation (i) and equation (ii), we get- \(\mathrm{m}_{1} \mathrm{~g}-\mathrm{m}_{2} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}+\mathrm{m}_{2} \mathrm{a}\) \(\mathrm{g}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)=\mathrm{a}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{a}=\frac{\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{g}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) By putting the value, \(\mathrm{a}=\frac{(5-4.8) \times 9.8}{5+4.8}\) \(\mathrm{a}=0.2 \mathrm{~ms}^{-2}\)
