363540
A block of mass \(M\) is pulled along a smooth horizontal surface with a rope of mass \(m\) by force \(F\). The acceleration of the block will be
1 \(\frac{F}{{\left( {M - m} \right)}}\)
2 \(\frac{F}{{\left( {M + m} \right)}}\)
3 \(\frac{F}{m}\)
4 \(\frac{F}{M}\)
Explanation:
When the block of mass \(M\) is pulled along smooth horizontal surface with rope of mass \(m\) by force \(F\), the total mass of the system is \((M + m)\). If the acceleration of the system is a, then \(F = (M + m).a\) \( \Rightarrow a = \frac{F}{{M + m}}\)
PHXI05:LAWS OF MOTION
363541
We can derive Newton’s
1 Second and third laws from the first law
2 First and second laws from the third law
3 Third and first laws from the second law
4 All the three laws are independent of each other
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363542
The mass of a lift is 2000 \(kg\). When the tension in the supporting cable is 28000 \(N\), then its accceleration is
1 \(30\,m{s^{ - 2}}\) downwards
2 \(14\,m{s^{ - 2}}\) upwards
3 \(4\,m{s^{ - 2}}\) downwards
4 \(4\,m{s^{ - 2}}\) upwards
Explanation:
Here, lift is accelerating upward at with acceleration \(a\). Hence, equation of motion is written as \(R - mg = ma\) \(28000 - 20000 = 2000\,a\) \(\left[ {\because \;g = 10\,m{s^{ - 2}}} \right]\) \( \Rightarrow \quad a = \frac{{8000}}{{2000}} = 4\,m{s^{ - 2}}\) upwards
PHXI05:LAWS OF MOTION
363543
Three forces \(F_{1}=10 N, F_{2}=8 N, F_{3}=6 N\) are acting on a particle of mass \(5\,kg\). The forces \(F_{2}\) and \(F_{3}\) are applied perpendicularly so that particle remains at rest. If the force \(F_{1}\) is removed, then the acceleration of the particle is
1 \(4.8\;m{s^{ - 2}}\)
2 \(2\;m{s^{ - 2}}\)
3 \(7\;m{s^{ - 2}}\)
4 \(0.5\;m{s^{ - 2}}\)
Explanation:
\(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=0\) If \(\vec{F}_{1}\) is removed, net force is given by \(\vec{F}=\sqrt{F_{2}^{2}+F_{3}^{2}+2 F_{2} F_{3} \cos 90^{\circ}}\) \( = \sqrt {64 + 36} = \sqrt {100} = 10\;N\) \(m = 5\;kg;a = \frac{F}{m} = \frac{{10}}{5} = 2\;m{s^{ - 2}}\)
JEE - 2023
PHXI05:LAWS OF MOTION
363544
Which one of the following statements is not true about Newton’s second law of motion \(\vec F = m\vec a\)?
1 The second law of motion is consistent with the first law.
2 The second law of motion is a vector law.
3 The second law of motion is applicable to a single point particle.
4 The second law of motion is not applicable for a system of particles.
363540
A block of mass \(M\) is pulled along a smooth horizontal surface with a rope of mass \(m\) by force \(F\). The acceleration of the block will be
1 \(\frac{F}{{\left( {M - m} \right)}}\)
2 \(\frac{F}{{\left( {M + m} \right)}}\)
3 \(\frac{F}{m}\)
4 \(\frac{F}{M}\)
Explanation:
When the block of mass \(M\) is pulled along smooth horizontal surface with rope of mass \(m\) by force \(F\), the total mass of the system is \((M + m)\). If the acceleration of the system is a, then \(F = (M + m).a\) \( \Rightarrow a = \frac{F}{{M + m}}\)
PHXI05:LAWS OF MOTION
363541
We can derive Newton’s
1 Second and third laws from the first law
2 First and second laws from the third law
3 Third and first laws from the second law
4 All the three laws are independent of each other
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363542
The mass of a lift is 2000 \(kg\). When the tension in the supporting cable is 28000 \(N\), then its accceleration is
1 \(30\,m{s^{ - 2}}\) downwards
2 \(14\,m{s^{ - 2}}\) upwards
3 \(4\,m{s^{ - 2}}\) downwards
4 \(4\,m{s^{ - 2}}\) upwards
Explanation:
Here, lift is accelerating upward at with acceleration \(a\). Hence, equation of motion is written as \(R - mg = ma\) \(28000 - 20000 = 2000\,a\) \(\left[ {\because \;g = 10\,m{s^{ - 2}}} \right]\) \( \Rightarrow \quad a = \frac{{8000}}{{2000}} = 4\,m{s^{ - 2}}\) upwards
PHXI05:LAWS OF MOTION
363543
Three forces \(F_{1}=10 N, F_{2}=8 N, F_{3}=6 N\) are acting on a particle of mass \(5\,kg\). The forces \(F_{2}\) and \(F_{3}\) are applied perpendicularly so that particle remains at rest. If the force \(F_{1}\) is removed, then the acceleration of the particle is
1 \(4.8\;m{s^{ - 2}}\)
2 \(2\;m{s^{ - 2}}\)
3 \(7\;m{s^{ - 2}}\)
4 \(0.5\;m{s^{ - 2}}\)
Explanation:
\(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=0\) If \(\vec{F}_{1}\) is removed, net force is given by \(\vec{F}=\sqrt{F_{2}^{2}+F_{3}^{2}+2 F_{2} F_{3} \cos 90^{\circ}}\) \( = \sqrt {64 + 36} = \sqrt {100} = 10\;N\) \(m = 5\;kg;a = \frac{F}{m} = \frac{{10}}{5} = 2\;m{s^{ - 2}}\)
JEE - 2023
PHXI05:LAWS OF MOTION
363544
Which one of the following statements is not true about Newton’s second law of motion \(\vec F = m\vec a\)?
1 The second law of motion is consistent with the first law.
2 The second law of motion is a vector law.
3 The second law of motion is applicable to a single point particle.
4 The second law of motion is not applicable for a system of particles.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI05:LAWS OF MOTION
363540
A block of mass \(M\) is pulled along a smooth horizontal surface with a rope of mass \(m\) by force \(F\). The acceleration of the block will be
1 \(\frac{F}{{\left( {M - m} \right)}}\)
2 \(\frac{F}{{\left( {M + m} \right)}}\)
3 \(\frac{F}{m}\)
4 \(\frac{F}{M}\)
Explanation:
When the block of mass \(M\) is pulled along smooth horizontal surface with rope of mass \(m\) by force \(F\), the total mass of the system is \((M + m)\). If the acceleration of the system is a, then \(F = (M + m).a\) \( \Rightarrow a = \frac{F}{{M + m}}\)
PHXI05:LAWS OF MOTION
363541
We can derive Newton’s
1 Second and third laws from the first law
2 First and second laws from the third law
3 Third and first laws from the second law
4 All the three laws are independent of each other
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363542
The mass of a lift is 2000 \(kg\). When the tension in the supporting cable is 28000 \(N\), then its accceleration is
1 \(30\,m{s^{ - 2}}\) downwards
2 \(14\,m{s^{ - 2}}\) upwards
3 \(4\,m{s^{ - 2}}\) downwards
4 \(4\,m{s^{ - 2}}\) upwards
Explanation:
Here, lift is accelerating upward at with acceleration \(a\). Hence, equation of motion is written as \(R - mg = ma\) \(28000 - 20000 = 2000\,a\) \(\left[ {\because \;g = 10\,m{s^{ - 2}}} \right]\) \( \Rightarrow \quad a = \frac{{8000}}{{2000}} = 4\,m{s^{ - 2}}\) upwards
PHXI05:LAWS OF MOTION
363543
Three forces \(F_{1}=10 N, F_{2}=8 N, F_{3}=6 N\) are acting on a particle of mass \(5\,kg\). The forces \(F_{2}\) and \(F_{3}\) are applied perpendicularly so that particle remains at rest. If the force \(F_{1}\) is removed, then the acceleration of the particle is
1 \(4.8\;m{s^{ - 2}}\)
2 \(2\;m{s^{ - 2}}\)
3 \(7\;m{s^{ - 2}}\)
4 \(0.5\;m{s^{ - 2}}\)
Explanation:
\(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=0\) If \(\vec{F}_{1}\) is removed, net force is given by \(\vec{F}=\sqrt{F_{2}^{2}+F_{3}^{2}+2 F_{2} F_{3} \cos 90^{\circ}}\) \( = \sqrt {64 + 36} = \sqrt {100} = 10\;N\) \(m = 5\;kg;a = \frac{F}{m} = \frac{{10}}{5} = 2\;m{s^{ - 2}}\)
JEE - 2023
PHXI05:LAWS OF MOTION
363544
Which one of the following statements is not true about Newton’s second law of motion \(\vec F = m\vec a\)?
1 The second law of motion is consistent with the first law.
2 The second law of motion is a vector law.
3 The second law of motion is applicable to a single point particle.
4 The second law of motion is not applicable for a system of particles.
363540
A block of mass \(M\) is pulled along a smooth horizontal surface with a rope of mass \(m\) by force \(F\). The acceleration of the block will be
1 \(\frac{F}{{\left( {M - m} \right)}}\)
2 \(\frac{F}{{\left( {M + m} \right)}}\)
3 \(\frac{F}{m}\)
4 \(\frac{F}{M}\)
Explanation:
When the block of mass \(M\) is pulled along smooth horizontal surface with rope of mass \(m\) by force \(F\), the total mass of the system is \((M + m)\). If the acceleration of the system is a, then \(F = (M + m).a\) \( \Rightarrow a = \frac{F}{{M + m}}\)
PHXI05:LAWS OF MOTION
363541
We can derive Newton’s
1 Second and third laws from the first law
2 First and second laws from the third law
3 Third and first laws from the second law
4 All the three laws are independent of each other
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363542
The mass of a lift is 2000 \(kg\). When the tension in the supporting cable is 28000 \(N\), then its accceleration is
1 \(30\,m{s^{ - 2}}\) downwards
2 \(14\,m{s^{ - 2}}\) upwards
3 \(4\,m{s^{ - 2}}\) downwards
4 \(4\,m{s^{ - 2}}\) upwards
Explanation:
Here, lift is accelerating upward at with acceleration \(a\). Hence, equation of motion is written as \(R - mg = ma\) \(28000 - 20000 = 2000\,a\) \(\left[ {\because \;g = 10\,m{s^{ - 2}}} \right]\) \( \Rightarrow \quad a = \frac{{8000}}{{2000}} = 4\,m{s^{ - 2}}\) upwards
PHXI05:LAWS OF MOTION
363543
Three forces \(F_{1}=10 N, F_{2}=8 N, F_{3}=6 N\) are acting on a particle of mass \(5\,kg\). The forces \(F_{2}\) and \(F_{3}\) are applied perpendicularly so that particle remains at rest. If the force \(F_{1}\) is removed, then the acceleration of the particle is
1 \(4.8\;m{s^{ - 2}}\)
2 \(2\;m{s^{ - 2}}\)
3 \(7\;m{s^{ - 2}}\)
4 \(0.5\;m{s^{ - 2}}\)
Explanation:
\(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=0\) If \(\vec{F}_{1}\) is removed, net force is given by \(\vec{F}=\sqrt{F_{2}^{2}+F_{3}^{2}+2 F_{2} F_{3} \cos 90^{\circ}}\) \( = \sqrt {64 + 36} = \sqrt {100} = 10\;N\) \(m = 5\;kg;a = \frac{F}{m} = \frac{{10}}{5} = 2\;m{s^{ - 2}}\)
JEE - 2023
PHXI05:LAWS OF MOTION
363544
Which one of the following statements is not true about Newton’s second law of motion \(\vec F = m\vec a\)?
1 The second law of motion is consistent with the first law.
2 The second law of motion is a vector law.
3 The second law of motion is applicable to a single point particle.
4 The second law of motion is not applicable for a system of particles.
363540
A block of mass \(M\) is pulled along a smooth horizontal surface with a rope of mass \(m\) by force \(F\). The acceleration of the block will be
1 \(\frac{F}{{\left( {M - m} \right)}}\)
2 \(\frac{F}{{\left( {M + m} \right)}}\)
3 \(\frac{F}{m}\)
4 \(\frac{F}{M}\)
Explanation:
When the block of mass \(M\) is pulled along smooth horizontal surface with rope of mass \(m\) by force \(F\), the total mass of the system is \((M + m)\). If the acceleration of the system is a, then \(F = (M + m).a\) \( \Rightarrow a = \frac{F}{{M + m}}\)
PHXI05:LAWS OF MOTION
363541
We can derive Newton’s
1 Second and third laws from the first law
2 First and second laws from the third law
3 Third and first laws from the second law
4 All the three laws are independent of each other
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363542
The mass of a lift is 2000 \(kg\). When the tension in the supporting cable is 28000 \(N\), then its accceleration is
1 \(30\,m{s^{ - 2}}\) downwards
2 \(14\,m{s^{ - 2}}\) upwards
3 \(4\,m{s^{ - 2}}\) downwards
4 \(4\,m{s^{ - 2}}\) upwards
Explanation:
Here, lift is accelerating upward at with acceleration \(a\). Hence, equation of motion is written as \(R - mg = ma\) \(28000 - 20000 = 2000\,a\) \(\left[ {\because \;g = 10\,m{s^{ - 2}}} \right]\) \( \Rightarrow \quad a = \frac{{8000}}{{2000}} = 4\,m{s^{ - 2}}\) upwards
PHXI05:LAWS OF MOTION
363543
Three forces \(F_{1}=10 N, F_{2}=8 N, F_{3}=6 N\) are acting on a particle of mass \(5\,kg\). The forces \(F_{2}\) and \(F_{3}\) are applied perpendicularly so that particle remains at rest. If the force \(F_{1}\) is removed, then the acceleration of the particle is
1 \(4.8\;m{s^{ - 2}}\)
2 \(2\;m{s^{ - 2}}\)
3 \(7\;m{s^{ - 2}}\)
4 \(0.5\;m{s^{ - 2}}\)
Explanation:
\(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=0\) If \(\vec{F}_{1}\) is removed, net force is given by \(\vec{F}=\sqrt{F_{2}^{2}+F_{3}^{2}+2 F_{2} F_{3} \cos 90^{\circ}}\) \( = \sqrt {64 + 36} = \sqrt {100} = 10\;N\) \(m = 5\;kg;a = \frac{F}{m} = \frac{{10}}{5} = 2\;m{s^{ - 2}}\)
JEE - 2023
PHXI05:LAWS OF MOTION
363544
Which one of the following statements is not true about Newton’s second law of motion \(\vec F = m\vec a\)?
1 The second law of motion is consistent with the first law.
2 The second law of motion is a vector law.
3 The second law of motion is applicable to a single point particle.
4 The second law of motion is not applicable for a system of particles.
