355342
The total mechanical energy of sun-planet system is:
1 \(+ve\)
2 \(-ve\)
3 Zero
4 Equal to its potential energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355343
Initially spring in its natural length, the block of mass \(0.25\;kg\) is released, then find out the value of maximum force by system on the floor
1 \(15\;N\)
2 \(20\;N\)
3 \(25\;N\)
4 \(30\;N\)
Explanation:
If \(x\) be the compression in the spring, when a block of \(0.25\;kg\) is released. From energy conservation, Potential energy of spring \(=\) Potential energy of block \( \Rightarrow \frac{1}{2}k{x^2} = mg\,x\) Where, \(k\) is force constant of spring. \(k x=2 m g=2 \times 0.25 g\) \(k x=0.5 g\) i.e. Force applied by the spring on the block, \(F=k x\) \(\Rightarrow F=0.5 \mathrm{~g}\) From Free body diagram, \(\therefore\) Maximum force by system on the floor, \(N=F+2 g\) \(\begin{aligned}& =0.5 g+2 g=2.5 g \quad\left(\because g=10 m^{-2}\right) \\& =25 N\end{aligned}\)
AIIMS - 2019
PHXI06:WORK ENERGY AND POWER
355344
A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
1 \(1 m\)
2 \(\sqrt{6} m\)
3 \(2 m\)
4 \(\sqrt{5} m\)
Explanation:
\(h = 10\sin 30 = 5\;m\) Let the maximum compression of spring be \(x\). Using the work-energy theorem, \(\Delta K=\) Work done by all forces \(\Delta K=m g h-f_{K} \times 2-\dfrac{1}{2} K x^{2}=0\) \(\left[\because K E_{i}=K E_{f}=0\right]\) Here \(f_{K}=\mu_{K} N=\mu_{K} m g=0.5 \times 5 \times 10=25 N\) \(5 \times 10 \times 5-25 \times 2-\dfrac{1}{2} \times 100 x^{2}=0\) On solving, we get \(x = 2{\rm{ }}m\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355345
A machine which is \(75 \%\) efficient, uses 12 \(J\) of energy in lifting 1 \(kg\) mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is
1 \(\sqrt {32} \,m{s^{ - 1}}\)
2 \(\sqrt {24} \,m{s^{ - 1}}\)
3 \(\sqrt {18} \,m{s^{ - 1}}\)
4 \(\sqrt {12} \,m{s^{ - 1}}\)
Explanation:
From energy conservation \(0.75(12J) = mgh = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {18} \;m/s\)
355342
The total mechanical energy of sun-planet system is:
1 \(+ve\)
2 \(-ve\)
3 Zero
4 Equal to its potential energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355343
Initially spring in its natural length, the block of mass \(0.25\;kg\) is released, then find out the value of maximum force by system on the floor
1 \(15\;N\)
2 \(20\;N\)
3 \(25\;N\)
4 \(30\;N\)
Explanation:
If \(x\) be the compression in the spring, when a block of \(0.25\;kg\) is released. From energy conservation, Potential energy of spring \(=\) Potential energy of block \( \Rightarrow \frac{1}{2}k{x^2} = mg\,x\) Where, \(k\) is force constant of spring. \(k x=2 m g=2 \times 0.25 g\) \(k x=0.5 g\) i.e. Force applied by the spring on the block, \(F=k x\) \(\Rightarrow F=0.5 \mathrm{~g}\) From Free body diagram, \(\therefore\) Maximum force by system on the floor, \(N=F+2 g\) \(\begin{aligned}& =0.5 g+2 g=2.5 g \quad\left(\because g=10 m^{-2}\right) \\& =25 N\end{aligned}\)
AIIMS - 2019
PHXI06:WORK ENERGY AND POWER
355344
A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
1 \(1 m\)
2 \(\sqrt{6} m\)
3 \(2 m\)
4 \(\sqrt{5} m\)
Explanation:
\(h = 10\sin 30 = 5\;m\) Let the maximum compression of spring be \(x\). Using the work-energy theorem, \(\Delta K=\) Work done by all forces \(\Delta K=m g h-f_{K} \times 2-\dfrac{1}{2} K x^{2}=0\) \(\left[\because K E_{i}=K E_{f}=0\right]\) Here \(f_{K}=\mu_{K} N=\mu_{K} m g=0.5 \times 5 \times 10=25 N\) \(5 \times 10 \times 5-25 \times 2-\dfrac{1}{2} \times 100 x^{2}=0\) On solving, we get \(x = 2{\rm{ }}m\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355345
A machine which is \(75 \%\) efficient, uses 12 \(J\) of energy in lifting 1 \(kg\) mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is
1 \(\sqrt {32} \,m{s^{ - 1}}\)
2 \(\sqrt {24} \,m{s^{ - 1}}\)
3 \(\sqrt {18} \,m{s^{ - 1}}\)
4 \(\sqrt {12} \,m{s^{ - 1}}\)
Explanation:
From energy conservation \(0.75(12J) = mgh = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {18} \;m/s\)
355342
The total mechanical energy of sun-planet system is:
1 \(+ve\)
2 \(-ve\)
3 Zero
4 Equal to its potential energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355343
Initially spring in its natural length, the block of mass \(0.25\;kg\) is released, then find out the value of maximum force by system on the floor
1 \(15\;N\)
2 \(20\;N\)
3 \(25\;N\)
4 \(30\;N\)
Explanation:
If \(x\) be the compression in the spring, when a block of \(0.25\;kg\) is released. From energy conservation, Potential energy of spring \(=\) Potential energy of block \( \Rightarrow \frac{1}{2}k{x^2} = mg\,x\) Where, \(k\) is force constant of spring. \(k x=2 m g=2 \times 0.25 g\) \(k x=0.5 g\) i.e. Force applied by the spring on the block, \(F=k x\) \(\Rightarrow F=0.5 \mathrm{~g}\) From Free body diagram, \(\therefore\) Maximum force by system on the floor, \(N=F+2 g\) \(\begin{aligned}& =0.5 g+2 g=2.5 g \quad\left(\because g=10 m^{-2}\right) \\& =25 N\end{aligned}\)
AIIMS - 2019
PHXI06:WORK ENERGY AND POWER
355344
A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
1 \(1 m\)
2 \(\sqrt{6} m\)
3 \(2 m\)
4 \(\sqrt{5} m\)
Explanation:
\(h = 10\sin 30 = 5\;m\) Let the maximum compression of spring be \(x\). Using the work-energy theorem, \(\Delta K=\) Work done by all forces \(\Delta K=m g h-f_{K} \times 2-\dfrac{1}{2} K x^{2}=0\) \(\left[\because K E_{i}=K E_{f}=0\right]\) Here \(f_{K}=\mu_{K} N=\mu_{K} m g=0.5 \times 5 \times 10=25 N\) \(5 \times 10 \times 5-25 \times 2-\dfrac{1}{2} \times 100 x^{2}=0\) On solving, we get \(x = 2{\rm{ }}m\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355345
A machine which is \(75 \%\) efficient, uses 12 \(J\) of energy in lifting 1 \(kg\) mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is
1 \(\sqrt {32} \,m{s^{ - 1}}\)
2 \(\sqrt {24} \,m{s^{ - 1}}\)
3 \(\sqrt {18} \,m{s^{ - 1}}\)
4 \(\sqrt {12} \,m{s^{ - 1}}\)
Explanation:
From energy conservation \(0.75(12J) = mgh = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {18} \;m/s\)
355342
The total mechanical energy of sun-planet system is:
1 \(+ve\)
2 \(-ve\)
3 Zero
4 Equal to its potential energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355343
Initially spring in its natural length, the block of mass \(0.25\;kg\) is released, then find out the value of maximum force by system on the floor
1 \(15\;N\)
2 \(20\;N\)
3 \(25\;N\)
4 \(30\;N\)
Explanation:
If \(x\) be the compression in the spring, when a block of \(0.25\;kg\) is released. From energy conservation, Potential energy of spring \(=\) Potential energy of block \( \Rightarrow \frac{1}{2}k{x^2} = mg\,x\) Where, \(k\) is force constant of spring. \(k x=2 m g=2 \times 0.25 g\) \(k x=0.5 g\) i.e. Force applied by the spring on the block, \(F=k x\) \(\Rightarrow F=0.5 \mathrm{~g}\) From Free body diagram, \(\therefore\) Maximum force by system on the floor, \(N=F+2 g\) \(\begin{aligned}& =0.5 g+2 g=2.5 g \quad\left(\because g=10 m^{-2}\right) \\& =25 N\end{aligned}\)
AIIMS - 2019
PHXI06:WORK ENERGY AND POWER
355344
A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
1 \(1 m\)
2 \(\sqrt{6} m\)
3 \(2 m\)
4 \(\sqrt{5} m\)
Explanation:
\(h = 10\sin 30 = 5\;m\) Let the maximum compression of spring be \(x\). Using the work-energy theorem, \(\Delta K=\) Work done by all forces \(\Delta K=m g h-f_{K} \times 2-\dfrac{1}{2} K x^{2}=0\) \(\left[\because K E_{i}=K E_{f}=0\right]\) Here \(f_{K}=\mu_{K} N=\mu_{K} m g=0.5 \times 5 \times 10=25 N\) \(5 \times 10 \times 5-25 \times 2-\dfrac{1}{2} \times 100 x^{2}=0\) On solving, we get \(x = 2{\rm{ }}m\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355345
A machine which is \(75 \%\) efficient, uses 12 \(J\) of energy in lifting 1 \(kg\) mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is
1 \(\sqrt {32} \,m{s^{ - 1}}\)
2 \(\sqrt {24} \,m{s^{ - 1}}\)
3 \(\sqrt {18} \,m{s^{ - 1}}\)
4 \(\sqrt {12} \,m{s^{ - 1}}\)
Explanation:
From energy conservation \(0.75(12J) = mgh = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {18} \;m/s\)
