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.
The image shows a graph of potential energy versus distance with marked regions AB, BC, CD, and DE.
### Related Formula
F = -fracdVdx$$F = -\frac{dV}{dx}$$
### 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|$|F| = |\text{slope of V-x graph}|$.
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$$F_{BC} > F_{AB} > F_{DE} > F_{CD}$$
### Pattern Recognition
In a V-x$V-x$ 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
Keywords:#potential energy graph#force from slope#V-x graph
More Work, Energy and Power Previous-Year Questions
Q10jee_main_2025_03_april_eveningCollisions and Spring Potential Energy
Consider two blocks A and B of masses m_1=10mathrm~kg$m_{1}=10\mathrm{~kg}$ and m_2=5mathrm~kg$m_{2}=5\mathrm{~kg}$ that are placed on a frictionless table. The block A moves with a constant speed v=3mathrm~m/s$v=3\mathrm{~m/s}$ towards the block B kept at rest. A spring with spring constant k=3000mathrm~N/m$k=3000\mathrm{~N/m}$ is attached with the block B as shown in the figure. Diagram showing Block A of mass m1 moving with velocity v towards Block B of mass m2 with a spring attached on a frictionless table. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is,
(Neglect the mass of the spring)
A. 0.2 m
B. 0.4 m
C. 0.1 m
D. 0.3 m
Solution
### Related Formula
By conservation of linear momentum, the common velocity v_textcm$v_{\text{cm}}$ of the combined mass system is:
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_textcm$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{\text{cm}}$$
By conservation of energy, the loss in Kinetic Energy during maximum compression converted to the potential energy of the spring:
Delta K = frac12 k x^2 Rightarrow frac12 m_1 v^2 - frac12 (m_1 + m_2) v_textcm^2 = frac12 k x^2$$\Delta K = \frac{1}{2} k x^2 \Rightarrow \frac{1}{2} m_1 v^2 - \frac{1}{2} (m_1 + m_2) v_{\text{cm}}^2 = \frac{1}{2} k x^2$$
### Core Logic
Given parameters:
- m_1 = 10mathrm~kg$m_1 = 10\mathrm{~kg}$, m_2 = 5mathrm~kg$m_2 = 5\mathrm{~kg}$
- Initial velocity of A: v = 3mathrm~m/s$v = 3\mathrm{~m/s}$
- Initial velocity of B: v_2 = 0$v_2 = 0$
- Spring constant k = 3000mathrm~N/m$k = 3000\mathrm{~N/m}$
### Step 1: Calculate the common center of mass velocity (v_textcm$v_{\text{cm}}$)
v_textcm = frac10 times 3 + 5 times 010 + 5 = frac3015 = 2mathrm~m/s$$v_{\text{cm}} = \frac{10 \times 3 + 5 \times 0}{10 + 5} = \frac{30}{15} = 2\mathrm{~m/s}$$
### Step 2: Apply Energy Conservation to find spring compression ($
### Step 2: Apply Energy Conservation to find spring compression ($x)
$)
$frac12 k x^2 = K_i - K_f$\frac{1}{2} k x^2 = K_i - K_f$$
$frac12 (3000) x^2 = left[ frac12 (10) (3^2) right] - left[ frac12 (10 + 5) (2^2) right]$\frac{1}{2} (3000) x^2 = \left[ \frac{1}{2} (10) (3^2) \right] - \left[ \frac{1}{2} (10 + 5) (2^2) \right]$$
$1500 x^2 = frac12(90) - frac12(15)(4)$1500 x^2 = \frac{1}{2}(90) - \frac{1}{2}(15)(4)$$
$1500 x^2 = 45 - 30 = 15$1500 x^2 = 45 - 30 = 15$$
$x^2 = frac151500 = frac1100$x^2 = \frac{15}{1500} = \frac{1}{100}$$
$x = frac110mathrm~m = 0.1mathrm~m$x = \frac{1}{10}\mathrm{~m} = 0.1\mathrm{~m}$
### Pattern Recognition
For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy.
$
### Pattern Recognition
For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy.
$frac12 mu v_textrel^2 = frac12 k x^2$\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{1}{2} k x^2$
where $
where $\mu = \frac{m_1 m_2}{m_1 + m_2} is the reduced mass. Here:
$ is the reduced mass. Here:
$mu = frac10 times 515 = frac103mathrm~kg$\mu = \frac{10 \times 5}{15} = \frac{10}{3}\mathrm{~kg}$$
$frac12 left(frac103right) (3)^2 = frac12 (3000) x^2 Rightarrow 15 = 1500 x^2 Rightarrow x = 0.1mathrm~m$\frac{1}{2} \left(\frac{10}{3}\right) (3)^2 = \frac{1}{2} (3000) x^2 \Rightarrow 15 = 1500 x^2 \Rightarrow x = 0.1\mathrm{~m}$$
Using reduced mass simplifies center-of-mass collision problems instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Work, Energy and Power
Q18jee_main_2025_03_april_eveningPower and Efficiency of Motor
A motor operating on 100mathrm~V$100\mathrm{~V}$ draws a current of 1 A. If the efficiency of the motor is 91.6\%$91.6\%$, then the loss of power in units of cal/s is :
A. 4
B. 8.4
C. 2
D. 6.2
Solution
### Related Formula
The electrical input power P_textin$P_{\text{in}}$ delivered to the motor is:
P_textin = V cdot I$$P_{\text{in}} = V \cdot I$$
The fraction of power lost is determined by the efficiency eta$\eta$:
P_textlost = P_textin - P_textout = P_textin (1 - eta)$$P_{\text{lost}} = P_{\text{in}} - P_{\text{out}} = P_{\text{in}} (1 - \eta)$$
To convert watts (mathrmJ/s$\mathrm{J/s}$) to calories per second:
P_textlost (mathrmcal/s) = fracP_textlost (mathrmW)4.2$$P_{\text{lost}} (\mathrm{cal/s}) = \frac{P_{\text{lost}} (\mathrm{W})}{4.2}$$
### Core Logic
Given parameters:
- Potential $
### Core Logic
Given parameters:
- Potential $V = 100\mathrm{~V}
- Current $
- Current $I = 1\mathrm{~A}
- Efficiency $
- Efficiency $\eta = 91.6\% = 0.916
### Step 1: Calculate Input Power ($
### Step 1: Calculate Input Power ($P_{\text{in}})
$)
$P_textin = 100 times 1 = 100mathrm~W$P_{\text{in}} = 100 \times 1 = 100\mathrm{~W}$
### Step 2: Calculate Power Loss in Watts
$
### Step 2: Calculate Power Loss in Watts
$P_textlost = 100 times (1 - 0.916) = 100 times 0.084 = 8.4mathrm~W$P_{\text{lost}} = 100 \times (1 - 0.916) = 100 \times 0.084 = 8.4\mathrm{~W}$
### Step 3: Convert Power Loss to cal/s
$
### Step 3: Convert Power Loss to cal/s
$P_textlost = frac8.44.2 = 2mathrm~cal/s$P_{\text{lost}} = \frac{8.4}{4.2} = 2\mathrm{~cal/s}$
### Pattern Recognition
Efficiency calculations often feature clean conversion rates. Recognizing that $
### Pattern Recognition
Efficiency calculations often feature clean conversion rates. Recognizing that $8.4 is exactly twice $ is exactly twice $4.2$ (the conversion factor for Joules to calories) simplifies the final step.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Work, Energy and Power
Q19jee_main_2025_03_april_eveningWork-Energy Theorem with Variable Force
A block of mass 1 kg, moving along x with speed v_i=10mathrm~m/s$v_{i}=10\mathrm{~m/s}$ enters a rough region ranging from x=0.1mathrm~m$x=0.1\mathrm{~m}$ to x=1.9mathrm~m$x=1.9\mathrm{~m}$. The retarding force acting on the block in this range is F_r=-kxmathrm~N$F_{r}=-kx\mathrm{~N}$ with k=10mathrm~N/m$k=10\mathrm{~N/m}$. Then the final speed of the block as it crosses rough region is :
A.10mathrm~m/s$10\mathrm{~m/s}$
B.4mathrm~m/s$4\mathrm{~m/s}$
C.6mathrm~m/s$6\mathrm{~m/s}$
D.8mathrm~m/s$8\mathrm{~m/s}$
Solution
### Related Formula
By the Work-Energy Theorem, the work done by the retarding force equals the change in kinetic energy:
W = Delta K = K_f - K_i$$W = \Delta K = K_f - K_i$$W = int_x_i^x_f F_r(x) dx = frac12 m v_f^2 - frac12 m v_i^2$$W = \int_{x_i}^{x_f} F_r(x) dx = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$
### Core Logic
Given parameters:
- Mass m = 1mathrm~kg$m = 1\mathrm{~kg}$
- Initial velocity v_i = 10mathrm~m/s$v_i = 10\mathrm{~m/s}$
- Region bounds: x_i = 0.1mathrm~m$x_i = 0.1\mathrm{~m}$, x_f = 1.9mathrm~m$x_f = 1.9\mathrm{~m}$
- Retarding force F_r = -kx = -10xmathrm~N$F_r = -kx = -10x\mathrm{~N}$
### Step 1: Calculate Work Done by the Retarding Force (W$W$)
W = int_0.1^1.9 (-10x) dx = -10 left[ fracx^22 right]_0.1^1.9 = -5 left[ (1.9)^2 - (0.1)^2 right]$$W = \int_{0.1}^{1.9} (-10x) dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9} = -5 \left[ (1.9)^2 - (0.1)^2 \right]$$
Using the algebraic identity a^2 - b^2 = (a-b)(a+b)$a^2 - b^2 = (a-b)(a+b)$:
(1.9)^2 - (0.1)^2 = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6$$(1.9)^2 - (0.1)^2 = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6$$W = -5 times 3.6 = -18mathrm~J$$W = -5 \times 3.6 = -18\mathrm{~J}$$
### Step 2: Solve for final velocity (v_f$v_f$)
Apply the Work-Energy Theorem:
-18 = frac12 (1) v_f^2 - frac12 (1) (10^2)$$-18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10^2)$$-18 = 0.5 v_f^2 - 50$$-18 = 0.5 v_f^2 - 50$$0.5 v_f^2 = 50 - 18 = 32$$0.5 v_f^2 = 50 - 18 = 32$$v_f^2 = 64 Rightarrow v_f = 8mathrm~m/s$$v_f^2 = 64 \Rightarrow v_f = 8\mathrm{~m/s}$$
### Pattern Recognition
Notice that integrating a linear force $
### Pattern Recognition
Notice that integrating a linear force $F = -kx yields a potential-energy-like term $ yields a potential-energy-like term $\frac{1}{2}k(x_f^2 - x_i^2). Combining this with the Work-Energy theorem gives $. Combining this with the Work-Energy theorem gives $\frac{1}{2} m v_f^2 + \frac{1}{2} k x_f^2 = \frac{1}{2} m v_i^2 + \frac{1}{2} k x_i^2$, which is identical to conservation of mechanical energy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Work, Energy and Power
Q18jee_main_2025_07_april_morningPower
An object of mass 1000 \, textg$1000 \, \text{g}$ experiences a time dependent force vecmathrmF = (2thatmathbfi + 3t^2hatmathbfj)mathrmN$\vec{\mathrm{F}} = (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}})\mathrm{N}$ . The power generated by the force at time t$t$ is:
### Related Formula
Instantaneous power P$P$ generated by a force is given by:
P = vecF cdot vecv$$P = \vec{F} \cdot \vec{v}$$
Newton's second law:
veca = fracvecFm = fracmathrmdvecvmathrmdt implies vecv = int veca \,mathrmdt$$\vec{a} = \frac{\vec{F}}{m} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \implies \vec{v} = \int \vec{a} \,\mathrm{d}t$$
### Core Logic
Convert mass to SI units:
m = 1000 mathrm~g = 1 mathrm~kg$$m = 1000 \mathrm{~g} = 1 \mathrm{~kg}$$
Calculate acceleration:
veca = frac2thatmathbfi + 3t^2hatmathbfj1 = 2thatmathbfi + 3t^2hatmathbfj$$\vec{a} = \frac{2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}}{1} = 2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}$$
### Step 1: Determine Velocity Vector
Assuming the object starts from rest at t = 0$t = 0$:
vecv = int_0^t (2thatmathbfi + 3t^2hatmathbfj) \,mathrmdt = t^2hatmathbfi + t^3hatmathbfj$$\vec{v} = \int_{0}^{t} (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \,\mathrm{d}t = t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}}$$
### Step 2: Calculate Power
Compute the dot product of force and velocity:
P = vecF cdot vecv = (2thatmathbfi + 3t^2hatmathbfj) cdot (t^2hatmathbfi + t^3hatmathbfj)$$P = \vec{F} \cdot \vec{v} = (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \cdot (t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}})$$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 mathrm~W$$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \mathrm{~W}$$
### Pattern Recognition
Sees: Time-dependent force $
### Pattern Recognition
Sees: Time-dependent force $\vec{F} \propto t^n on a 1 kg mass.
Shortcut: For $ on a 1 kg mass.
Shortcut: For $m=1 kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $ kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $\int at^n \mathrm{d}t = \frac{a}{n+1} t^{n+1}, power component becomes $, power component becomes $\frac{a^2}{n+1} t^{2n+1}. Here, $. Here, $2^2/2 t^3 + 3^2/3 t^5 = 2t^3 + 3t^5 \mathrm{~W}$.
### 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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