Top 50 Most Repeated WORK ENERGY AND POWER PYQs | JEE MAINS
A curated collection of the most important questions from WORK ENERGY AND POWER, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from WORK ENERGY AND POWER, fully solved with step-by-step concepts to prepare for JEE MAINS.
Average power \(P_{avg}=\Delta K/\Delta t\). The kinetic energy increase \(\Delta K = 80 J‑20 J = 60 J\). Over 4 s, \(P_{avg}=60/4 = 15 W\)....
Read Full Step-by-Step Solution →Use $P = \frac{dW}{dt}$, integrate to find work, apply work-energy theorem....
Read Full Step-by-Step Solution →For a one‑dimensional elastic collision, \(v_{1}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\,u_{1}\). Substituting \(m_{1}=m\) and \(m_{2}=2m\) gives \(v_{1}=\fr...
Read Full Step-by-Step Solution →Work done by gravity is negative as force opposes displacement....
Read Full Step-by-Step Solution →Elastic collision implies $e = 1$, relative speed after = relative speed before....
Read Full Step-by-Step Solution →Work by force = 40 J, friction = 16 J, net = 24 J....
Read Full Step-by-Step Solution →Work done = friction force × displacement. $f_k = \mu mg = 0.3 \times 20 = 6 N$, $W = 6 \times 5 = 30 J$....
Read Full Step-by-Step Solution →$W = \Delta KE = \frac{1}{2} \times 4 \times (20^2 - 10^2) = 600\,J$....
Read Full Step-by-Step Solution →For a perfectly inelastic collision, the centre‑of‑mass velocity is conserved. Total momentum before collision is
Using conservation of mechanical energy: loss in potential energy = gain in kinetic energy. So, $mgh = \frac{1}{2}mv^2$. Solving, $v = \sqrt{2gh} = \s...
Read Full Step-by-Step Solution →Center of mass velocity remains unchanged in absence of external forces....
Read Full Step-by-Step Solution →First compute the acceleration $a=F/m=4\,\text{m/s}^{2}$. Using $s=\frac{1}{2}at^{2}$, the time to travel 3 m is $t=\sqrt{2s/a}=\sqrt{1.5}\approx1.225...
Read Full Step-by-Step Solution →By conservation: $mgh = \frac{1}{2}mv^2$ → $v = \sqrt{2gh} = \sqrt{100} = 10\,\text{m/s}$ (g=10)....
Read Full Step-by-Step Solution →Foundational check for Work, Energy and Power. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Average power \(P_{avg}=\Delta K/\Delta t\). The kinetic energy increase \(\Delta K = 80 J‑20 J = 60 J\). Over 4 s, \(P_{avg}=60/4 = 15 W\)....
Read Full Step-by-Step Solution →Use $P = \frac{dW}{dt}$, integrate to find work, apply work-energy theorem....
Read Full Step-by-Step Solution →For a one‑dimensional elastic collision, \(v_{1}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\,u_{1}\). Substituting \(m_{1}=m\) and \(m_{2}=2m\) gives \(v_{1}=\fr...
Read Full Step-by-Step Solution →Work done by gravity is negative as force opposes displacement....
Read Full Step-by-Step Solution →Elastic collision implies $e = 1$, relative speed after = relative speed before....
Read Full Step-by-Step Solution →Work by force = 40 J, friction = 16 J, net = 24 J....
Read Full Step-by-Step Solution →Work done = friction force × displacement. $f_k = \mu mg = 0.3 \times 20 = 6 N$, $W = 6 \times 5 = 30 J$....
Read Full Step-by-Step Solution →$W = \Delta KE = \frac{1}{2} \times 4 \times (20^2 - 10^2) = 600\,J$....
Read Full Step-by-Step Solution →For a perfectly inelastic collision, the centre‑of‑mass velocity is conserved. Total momentum before collision is
Using conservation of mechanical energy: loss in potential energy = gain in kinetic energy. So, $mgh = \frac{1}{2}mv^2$. Solving, $v = \sqrt{2gh} = \s...
Read Full Step-by-Step Solution →Center of mass velocity remains unchanged in absence of external forces....
Read Full Step-by-Step Solution →First compute the acceleration $a=F/m=4\,\text{m/s}^{2}$. Using $s=\frac{1}{2}at^{2}$, the time to travel 3 m is $t=\sqrt{2s/a}=\sqrt{1.5}\approx1.225...
Read Full Step-by-Step Solution →By conservation: $mgh = \frac{1}{2}mv^2$ → $v = \sqrt{2gh} = \sqrt{100} = 10\,\text{m/s}$ (g=10)....
Read Full Step-by-Step Solution →Foundational check for Work, Energy and Power. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Average power \(P_{avg}=\Delta K/\Delta t\). The kinetic energy increase \(\Delta K = 80 J‑20 J = 60 J\). Over 4 s, \(P_{avg}=60/4 = 15 W\)....
Read Full Step-by-Step Solution →Use $P = \frac{dW}{dt}$, integrate to find work, apply work-energy theorem....
Read Full Step-by-Step Solution →For a one‑dimensional elastic collision, \(v_{1}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\,u_{1}\). Substituting \(m_{1}=m\) and \(m_{2}=2m\) gives \(v_{1}=\fr...
Read Full Step-by-Step Solution →Work done by gravity is negative as force opposes displacement....
Read Full Step-by-Step Solution →Elastic collision implies $e = 1$, relative speed after = relative speed before....
Read Full Step-by-Step Solution →Work by force = 40 J, friction = 16 J, net = 24 J....
Read Full Step-by-Step Solution →Work done = friction force × displacement. $f_k = \mu mg = 0.3 \times 20 = 6 N$, $W = 6 \times 5 = 30 J$....
Read Full Step-by-Step Solution →$W = \Delta KE = \frac{1}{2} \times 4 \times (20^2 - 10^2) = 600\,J$....
Read Full Step-by-Step Solution →For a perfectly inelastic collision, the centre‑of‑mass velocity is conserved. Total momentum before collision is
Using conservation of mechanical energy: loss in potential energy = gain in kinetic energy. So, $mgh = \frac{1}{2}mv^2$. Solving, $v = \sqrt{2gh} = \s...
Read Full Step-by-Step Solution →Center of mass velocity remains unchanged in absence of external forces....
Read Full Step-by-Step Solution →First compute the acceleration $a=F/m=4\,\text{m/s}^{2}$. Using $s=\frac{1}{2}at^{2}$, the time to travel 3 m is $t=\sqrt{2s/a}=\sqrt{1.5}\approx1.225...
Read Full Step-by-Step Solution →By conservation: $mgh = \frac{1}{2}mv^2$ → $v = \sqrt{2gh} = \sqrt{100} = 10\,\text{m/s}$ (g=10)....
Read Full Step-by-Step Solution →Foundational check for Work, Energy and Power. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Average power \(P_{avg}=\Delta K/\Delta t\). The kinetic energy increase \(\Delta K = 80 J‑20 J = 60 J\). Over 4 s, \(P_{avg}=60/4 = 15 W\)....
Read Full Step-by-Step Solution →Use $P = \frac{dW}{dt}$, integrate to find work, apply work-energy theorem....
Read Full Step-by-Step Solution →For a one‑dimensional elastic collision, \(v_{1}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\,u_{1}\). Substituting \(m_{1}=m\) and \(m_{2}=2m\) gives \(v_{1}=\fr...
Read Full Step-by-Step Solution →Work done by gravity is negative as force opposes displacement....
Read Full Step-by-Step Solution →Elastic collision implies $e = 1$, relative speed after = relative speed before....
Read Full Step-by-Step Solution →