A bead of mass m slides without friction on the wall of a vertical circular hoop of radius R as shown in figure. The…

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More Work, Energy and Power Previous-Year Questions — Page 3

Q19 jee_main_2025_03_april_evening Work-Energy Theorem with Variable Force

A block of mass 1 kg, moving along x with speed

vᵢ=10~m/s

enters a rough region ranging from

x=0.1~m

x=1.9~m

. The retarding force acting on the block in this range is

Fᵣ=-kx~N

with

k=10~N/m

. Then the final speed of the block as it crosses rough region is :

A. $10~m/s$ $10\mathrm{~m/s}$
B. $4~m/s$ $4\mathrm{~m/s}$
C. $6~m/s$ $6\mathrm{~m/s}$
D. $8~m/s$ $8\mathrm{~m/s}$

Solution

Related Formula

By the Work-Energy Theorem, the work done by the retarding force equals the change in kinetic energy:

W = Δ K = Kf - Kᵢ

W = \intxᵢxf Fᵣ(x) dx = (1)/(2) m vf² - (1)/(2) m vᵢ²

Core Logic

Given parameters:

Mass $m = 1~kg$ $m = 1\mathrm{~kg}$
Initial velocity $vᵢ = 10~m/s$ $v_i = 10\mathrm{~m/s}$
Region bounds: $xᵢ = 0.1~m$ $x_i = 0.1\mathrm{~m}$, $xf = 1.9~m$ $x_f = 1.9\mathrm{~m}$
Retarding force $Fᵣ = -kx = -10x~N$ $F_r = -kx = -10x\mathrm{~N}$

Step 1: Calculate Work Done by the Retarding Force ( $W$ $W$)

W = \int0.11.9 (-10x) dx = -10 [ (x²)/(2) ]0.11.9 = -5 [ (1.9)² - (0.1)² ]

Using the algebraic identity $a² - b² = (a-b)(a+b)$ $a^2 - b^2 = (a-b)(a+b)$:

(1.9)² - (0.1)² = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6

W = -5 \times 3.6 = -18~J

Step 2: Solve for final velocity ( $vf$ $v_f$)

Apply the Work-Energy Theorem:

-18 = (1)/(2) (1) vf² - (1)/(2) (1) (10²)

-18 = 0.5 vf² - 50

0.5 vf² = 50 - 18 = 32

vf² = 64 \Rightarrow vf = 8~m/s

Pattern Recognition Notice that integrating a linear force$ Pattern Recognition Notice that integrating a linear force $F = -kxyields a potential-energy-like term$ 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. Chapter Mix Class 11 Physics: Work, Energy and Power

Q18 jee_main_2025_07_april_morning Power

An object of mass

1000 g

experiences a time dependent force

F = (2t i + 3t² j)N

. The power generated by the force at time

t

is:

A. $(2t² + 3t³)W$ $(2t^{2} + 3t^{3})\mathrm{W}$
B. $(2t² + 18t³)W$
C. $(3t³ + 5t⁵)W$ $(3t^{3} + 5t^{5})\mathrm{W}$
D. $(2t³ + 3t⁵)W$ $(2t^{3} + 3t^{5})\mathrm{W}$

Solution

Related Formula

Instantaneous power $P$ $P$ generated by a force is given by:

P = F \cdot v

Newton's second law:

a = Fm = d vdt v = \int a dt

Core Logic

Convert mass to SI units:

m = 1000 ~g = 1 ~kg

Calculate acceleration:

a = 2t i + 3t² j1 = 2t i + 3t² j

Step 1: Determine Velocity Vector

Assuming the object starts from rest at $t = 0$ $t = 0$:

v = \int₀t (2t i + 3t² j) dt = t² i + t³ j

Step 2: Calculate Power

Compute the dot product of force and velocity:

P = F \cdot v = (2t i + 3t² j) \cdot (t² i + t³ j)

P = (2t)(t²) + (3t²)(t³) = 2t³ + 3t⁵ ~W

Pattern Recognition Sees: Time-dependent force$ Pattern Recognition Sees: Time-dependent force $\vec{F} \propto t^non a 1 kg mass. Shortcut: For$ on a 1 kg mass. Shortcut: For $m=1kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since$ 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, $2^2/2 t^3 + 3^2/3 t^5 = 2t^3 + 3t^5 \mathrm{~W}$. Chapter Mix Class 11 Physics: Work, Energy and Power

Q jee_main_2025_08_april_evening Conservation of Mechanical Energy

A block of mass

2~kg

is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is

2~m

and spring constant is

200~N/m

. The block is pushed such that the length of the spring becomes

1~m

and then released. At distance

x~m

(

x < 2

) from the wall, the speed of the block will be:

A. $10[1 - (2 - x)]3/2~m/s$ $10[1 - (2 - x)]^{3/2}\mathrm{~m/s}$
B. $10[1 - (2 - x)² ]1/2~m/s$ $10[1 - (2 - x)^2 ]^{1/2}\mathrm{~m/s}$
C. $10[1 - (2 - x)²]~m/s$ $10[1 - (2 - x)^{2}]\mathrm{~m/s}$
D. $10[1 - (2 - x)²]²~m/s$ $10[1 - (2 - x)^{2}]^{2}\mathrm{~m/s}$

Solution

Related Formula

E = Kᵢ + Uᵢ = Kf + Uf

U = (1)/(2) k y²

where, $E$ $E$ = total mechanical energy $K = (1)/(2) m v²$ $K = \frac{1}{2} m v^2$ = kinetic energy $U$ $U$ = potential energy of spring with deformation $y$ $y$

Core Logic

Given parameters:

Mass, $m = 2~kg$ $m = 2\mathrm{~kg}$
Natural length of spring, $L₀ = 2~m$ $L_0 = 2\mathrm{~m}$
Spring constant, $k = 200~N/m$ $k = 200\mathrm{~N/m}$

Initial State (when block is pushed):

Length of spring is $1~m$ $1\mathrm{~m}$.
Deformation (compression), $yᵢ = L₀ - 1 = 2 - 1 = 1~m$ $y_i = L_0 - 1 = 2 - 1 = 1\mathrm{~m}$.
Released from rest: $vᵢ = 0 Kᵢ = 0$ $v_i = 0 \implies K_i = 0$.

x

Since the spring is attached to the wall, its length is $x~m$ $x\mathrm{~m}$.
Deformation (compression) at this position, $yf = L₀ - x = (2 - x)~m$ $y_f = L_0 - x = (2 - x)\mathrm{~m}$.
Kinetic energy $Kf = (1)/(2) m v² = (1)/(2) (2) v² = v²$ .

Step 1: Conservation of Energy Equation

Equate initial and final energies:

Kᵢ + Uᵢ = Kf + Uf

0 + (1)/(2) k yᵢ² = (1)/(2) m v² + (1)/(2) k yf²

Substitute the parameters:

(1)/(2) (200) (1)² = v² + (1)/(2) (200) (2 - x)²

100 = v² + 100 (2 - x)²

v² = 100 [ 1 - (2 - x)² ]

v = 10 [ 1 - (2 - x)² ]1/2~m/s

Pattern Recognition

Sees: Horizontal spring-mass energy conservation. Trap: The deformation is not $x$ $x$; it is the difference from natural length, i.e., $(L₀ - x) = (2 - x)$ $(L_0 - x) = (2 - x)$. Shortcut: Writing out energy conservation directly allows mass to cancel beautifully, simplifying the algebra immediately. ✓

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Oscillations

Q11 jee_main_2025_29_jan_evening Elastic Collisions in One Dimension

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities

vA = 5~m/s, vB = 2~m/s, vC = 4~m/s

. If we wait sufficiently long for elastic collision to happen, then

vA = 4~m/s, vB = 2~m/s, vC = 5~m/s

will be the final velocities. Reason (R): In an elastic collision between identical masses, two objects exchange their velocities. In the light of the above statements, choose the correct answer from the options given below:

A. $Both (A) and (R) are true but (R) is NOT the correct explanation of (A)$
B. $(A) is true but (R) is false$ $\text{(A) is true but (R) is false}$
C. $Both (A) and (R) are true and (R) is the correct explanation of (A)$
D. $(A) is false but (R) is true$ $\text{(A) is false but (R) is true}$

Solution

Related Formula

v₁' = v₂ and v₂' = v₁

for perfectly elastic collision ( $e=1$ $e=1$) when masses are identical ( $m₁ = m₂$ $m_1 = m_2$).

Core Logic

Reason (R) states that identical masses exchange their velocities during elastic collision, which is mathematically correct.

Let us trace the sequence of collisions chronologically:

Since $vA = 5~m/s$ $v_A = 5\mathrm{~m/s}$ and $vB = 2~m/s$ $v_B = 2\mathrm{~m/s}$, sphere A collides with sphere B. After this collision, they swap velocities:

vA' = 2~m/s, vB' = 5~m/s

Now sphere B has velocity $vB' = 5~m/s$ $v_B' = 5\mathrm{~m/s}$ and sphere C has $vC = 4~m/s$ $v_C = 4\mathrm{~m/s}$. Sphere B will collide with C. After swapping:

vB'' = 4~m/s, vC' = 5~m/s

Looking at the values now: $vA' = 2~m/s$ $v_A' = 2\mathrm{~m/s}$ and $vB'' = 4~m/s$ $v_B'' = 4\mathrm{~m/s}$. No more collisions occur.

Therefore, the final velocities are $vA = 2~m/s, vB = 4~m/s, vC = 5~m/s$ . The values given in Assertion (A) are wrong. Hence, (A) is false but (R) is true.

Elastic Collisions Step 1 diagram for Q11 - JEE Main 2025 Evening

Pattern Recognition

Velocity exchange happens pairwise in sequential order. Do not try to solve simultaneous conservation laws across all three blocks at once; handle each collision step-by-step from left to right.

Chapter Mix

Class 11 Physics: Work, Energy and Power

Solution & Explanation

Core Logic

Step 1: Final Kinematic Value

Pattern Recognition

Chapter Mix

More Work, Energy and Power Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Work Done by the Retarding Force (W$W$)

Step 2: Solve for final velocity (vf$v_f$)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Velocity Vector

Step 2: Calculate Power

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conservation of Energy Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Work, Energy and Power Questions — jee_main_2025_28_jan_morning

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Step 1: Calculate Work Done by the Retarding Force ( $W$ $W$)

Step 2: Solve for final velocity ( $vf$ $v_f$)