Potential energy (V) versus distance (x) is given by the graph, Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.
Potential Energy diagram for Q26 - JEE Main 2026 Morning
The image shows a graph of potential energy versus distance with marked regions AB, BC, CD, and DE.

Solution & Explanation

### Related Formula F = -fracdVdx ### Core Logic The magnitude of the force is given by the absolute value of the slope of the potential energy versus position curve: |F| = |textslope of V-x graph|. Comparing the slopes of different regions from the graph: - Region BC has the steepest slope (largest magnitude). - Region AB has a moderate positive slope. - Region DE has a moderate negative slope, but less steep than AB. - Region CD is perfectly flat (slope = 0). ### Step 1: Ranking Forces Therefore, magnitudes of the force follow the order: F_BC > F_AB > F_DE > F_CD ### Pattern Recognition In a V-x graph, steeper slopes always mean a stronger restoring or repulsive force. A flat line means zero force (equilibrium). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

More Work, Energy and Power Previous-Year Questions — Page 4

Q jee_main_2025_29_jan_morning Vertical Circular Motion
A body of mass 'm' connected to a massless and unstretchable string goes in a vertical circle of radius 'R' under gravity g. The other end of the string is fixed at the center of a circle. If velocity at top of circular path is n sqrtgR , where, n geq 1 , then ratio of kinetic energy of the body at bottom to that at top of the circle is
  • A. fracmathrmnmathrmn + 4
  • B. fracmathrmn + 4mathrmn
  • C. fracmathrmn^2mathrmn^2 + 4
  • D. fracmathrmn^2 + 4mathrmn^2

Solution

### Related Formula V_textBottom = sqrtV_textTop^2 + 4gR ### Core Logic Given velocity at the top position : V_textTop = sqrtn^2 gR By work-energy theorem, mechanical energy conservation between the top and bottom positions gives : V_textBottom = sqrtn^2 gR + 4gR Since KE = frac12m v^2, the ratio of kinetic energy at the bottom to that at the top is textRatio = fracV_textBottom^2V_textTop^2 = fracn^2 gR + 4gRn^2 gR = fracn^2 + 4n^2 ### Pattern Recognition Kinetic energy change in vertical circles always gains a fixed additive value of 2mg(2R) = 4mgR due to gravity work. ### Chapter Mix Class 11 Physics: Work, Energy and Power
Q46 jee_main_2024_01_february_morning Collisions
A simple pendulum of length 1mathrm~m has a wooden bob of mass 1mathrm~kg. It is struck by a bullet of mass 10^-2mathrm~kg moving with a speed of 2 times 10^2mathrm~ms^-1. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is (use g = 10mathrm~m/s^2):
  • A. 0.30 m
  • B. 0.20 m
  • C. 0.35 m
  • D. 0.40 m

Solution

### Related Formula Conservation of Linear Momentum during embedded collision: mu = (M + m)V Conservation of Mechanical Energy during vertical rise: frac12(M+m)V^2 = (M+m)gh implies h = fracV^22g ### Core Logic Given data: m = 10^-2mathrm~kg, u = 2 times 10^2mathrm~ms^-1, M = 1mathrm~kg, g = 10mathrm~m/s^2. Apply momentum balance: 10^-2 times (2 times 10^2) = (1 + 0.01)V 2 = 1.01V implies V approx 2mathrm~ms^-1\ text(since 1.01 approx 1text) ### Step 1: Compute Maximum Rise Height Using the work-energy relation for the subsequent upward swing: h = fracV^22g = frac2^22 times 10 = frac420 = 0.20mathrm~m ### Pattern Recognition Perfectly inelastic collision approximation: because m ll M, we can simplify M+m approx M during velocity matching to complete the calculation rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power Class 11 Physics: Laws of Motion
Q37 jee_main_2024_29_january_evening Conservation of Mechanical Energy
The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10text m. If it dissipates 10\% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use g = 10text m s^-2]
  • A. 6sqrt5text m s^-1
  • B. 5sqrt6text m s^-1
  • C. 5sqrt5text m s^-1
  • D. 2sqrt5text m s^-1

Solution

### Related Formula Potential energy at the horizontal release point relative to the lowest point: U_i = mgell If 10\% of energy is dissipated, the remaining 90\% is converted entirely to kinetic energy at the lowest point: E_f = 0.90 times U_i = frac12mv^2 ### Core Logic Let the horizontal position be our reference for potential energy relative to the lowest point. * Initial potential energy: U_i = mgell * Energy remaining after 10\% loss: 0.90(mgell) Equating this to kinetic energy at the bottom: frac910 mgell = frac12 mv^2 ### Step 1: Calculate the Velocity We can cancel m from both sides: frac910 gell = frac12 v^2 Substitute ell = 10text m and g = 10text m s^-2: frac910 (10)(10) = frac12 v^2 90 = frac12 v^2 implies v^2 = 180 v = sqrt180 = sqrt36 times 5 = 6sqrt5text m s^-1
Pendulum trajectory diagram for Q37
Pendulum trajectory diagram for Q37
### Pattern Recognition Whenever there is a fractional energy loss x, use the formula: v = sqrt2(1-x)gell. Substituting x = 0.1 gives v = sqrt1.8gell directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power
Q42 jee_main_2024_29_january_evening Vertical Circular Motion
A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies frac(mathrmK.E.)_mathrmA(mathrmK.E.)_mathrmB is:
Vertical circular motion path of a pendulum bob for Q42 - JEE Main 2024 29 January Shift 2
The diagram displays a bob of mass m in vertical circular motion with velocity indicators at points A, B, and C.
  • A. 3:2
  • B. 5:1
  • C. 2:5
  • D. 1:5

Solution

### Related Formula For a body to just complete a vertical loop of radius L: * Speed at the lowest point A: v_A = sqrt5gL * Speed at the highest point B: v_B = sqrtgL ### Core Logic The kinetic energy at any point is given by: textK.E. = frac12mv^2 Thus: (textK.E.)_A = frac12m v_A^2 = frac12m(5gL) (textK.E.)_B = frac12m v_B^2 = frac12m(gL) ### Step 1: Calculate the Ratio Taking the ratio of the kinetic energies at A and B: frac(textK.E.)_A(textK.E.)_B = fracfrac12m(5gL)frac12m(gL) = frac51 = 5:1 ### Pattern Recognition Since textK.E. propto v^2, the ratio of kinetic energies is simply the ratio of the squares of the critical velocities at the bottom and top of the vertical loop: (sqrt5gL)^2 : (sqrtgL)^2 = 5:1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power
Q39 jee_main_2024_27_jan_morning Kinetic Energy and Momentum
Two bodies of mass 4text g and 25text g are moving with equal kinetic energies. The ratio of the magnitude of their linear momentum is:
  • A. 3:5
  • B. 5:4
  • C. 2:5
  • D. 4:5

Solution

### Related Formula K = fracP^22m implies P = sqrt2mK Where P is linear momentum, m is mass, and K is kinetic energy. ### Core Logic Given K_1 = K_2, the momentum ratio simplifies directly to the square root of their masses: fracP_1P_2 = sqrtfracm_1m_2 ### Step 1: Calculate the value Substitute m_1 = 4text g and m_2 = 25text g: fracP_1P_2 = sqrtfrac425 = frac25 ### Pattern Recognition For constant kinetic energy tracking profiles, momentum maps proportionally to sqrtm. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

More Work, Energy and Power Questions — jee_main_2026_21_jan_morning

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