A 4 kg mass moves under the influence of a forcevecmathrmF = (4mathrmt^3hatmathrmi - 3mathrmthatmathrmj)text N$\vec{\mathrm{F}} = (4\mathrm{t}^3\hat{\mathrm{i}} - 3\mathrm{t}\hat{\mathrm{j}})\text{ N}$ where t is the time in second. If mass starts from origin at t = 0$t = 0$, the velocity and position after t = 2texts$t = 2\text{s}$ will be:
Keywords:#A 4 kg mass moves under the influence of a force#JEE Main 2026 Morning Q31#Laws of Motion JEE Main 2026#Force and Acceleration JEE Main 2026
More Laws of Motion Previous-Year Questions
Q16jee_main_2025_02_april_eveningEquilibrium of Forces
A body of mass 1mathrmkg$1\mathrm{kg}$ is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions mathbfT_1$\mathbf{T}_1$ and mathbfT_2$\mathbf{T}_2$ , respectively, are (in N):
The diagram shows a suspended mass of 1 kg held by two strings making angles of 60 and 30 degrees with the horizontal.
A.5, 5sqrt3$5, 5\sqrt{3}$
B.5sqrt3, 5$5\sqrt{3}, 5$
C.5sqrt3, 5sqrt3$5\sqrt{3}, 5\sqrt{3}$
D.5, 5$5, 5$
Solution
### Related Formula
For a system in static equilibrium:
sum F_x = 0 quad textand quad sum F_y = 0$$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$$
### Core Logic
Let's resolve the tension forces vecT_1$\vec{T}_1$ and vecT_2$\vec{T}_2$ into horizontal and vertical components:
- T_1$T_1$ makes 60^circ$60^\circ$ with the horizontal.
- T_2$T_2$ makes 30^circ$30^\circ$ with the horizontal.
- Downward gravitational force: W = m g = 1 times 10 = 10 \ mathrmN$W = m g = 1 \times 10 = 10 \ \mathrm{N}$.
1. **Horizontal Equilibrium (sum F_x = 0$\sum F_x = 0$):**
T_1 cos 60^circ = T_2 cos 30^circ$$T_1 \cos 60^\circ = T_2 \cos 30^\circ$$T_1 cdot frac12 = T_2 cdot fracsqrt32 implies T_1 = T_2 sqrt3$$T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{3}}{2} \implies T_1 = T_2 \sqrt{3}$$
2. **Vertical Equilibrium (sum F_y = 0$\sum F_y = 0$):**
T_1 sin 60^circ + T_2 sin 30^circ = m g = 10$$T_1 \sin 60^\circ + T_2 \sin 30^\circ = m g = 10$$T_1 cdot fracsqrt32 + T_2 cdot frac12 = 10$$T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 10$$
### Step 1: Solve for Tensions
Substitute T_1 = T_2 sqrt3$T_1 = T_2 \sqrt{3}$ into the vertical equilibrium equation:
(T_2 sqrt3) fracsqrt32 + fracT_22 = 10$$(T_2 \sqrt{3}) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10$$frac3 T_22 + fracT_22 = 10 implies 2 T_2 = 10 implies T_2 = 5 \ mathrmN$$\frac{3 T_2}{2} + \frac{T_2}{2} = 10 \implies 2 T_2 = 10 \implies T_2 = 5 \ \mathrm{N}$$
Substitute T_2$T_2$ back to obtain T_1$T_1$:
T_1 = 5 sqrt3 \ mathrmN$$T_1 = 5 \sqrt{3} \ \mathrm{N}$$
Thus, the tension magnitudes are T_1 = 5sqrt3 \ mathrmN$T_1 = 5\sqrt{3} \ \mathrm{N}$ and T_2 = 5 \ mathrmN$T_2 = 5 \ \mathrm{N}$.
### Pattern Recognition
Sees: Suspending particle static equilibrium with asymmetric strings.
Trap: Associating components with incorrect trigonometry axes or swapping T_1$T_1$ and T_2$T_2$ in options.
Shortcut: Since the incline of T_1$T_1$ (60^circ$60^\circ$) is steeper than that of T_2$T_2$ (30^circ$30^\circ$), T_1$T_1$ must carry a larger portion of the load, meaning T_1 > T_2$T_1 > T_2$. From the choices, only (2) satisfies this hierarchy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q14jee_main_2025_07_april_morningFriction
A cubic block of mass m is sliding down on an inclined plane at 60^circ$60^{\circ}$ with an acceleration of fracg2$\frac{g}{2}$ , the value of coefficient of kinetic friction is
A.sqrt3 - 1$\sqrt{3} - 1$
B.fracsqrt32$\frac{\sqrt{3}}{2}$
C.fracsqrt23$\frac{\sqrt{2}}{3}$
D.1 - fracsqrt32$1 - \frac{\sqrt{3}}{2}$
Solution
### Related Formula
For a block sliding down an inclined plane of inclination theta$\theta$:
mg sintheta - f_k = ma$$mg \sin\theta - f_k = ma$$
Where normal reaction is N = mg costheta$N = mg \cos\theta$ and kinetic friction is:
f_k = mu_k N = mu_k mg costheta$$f_k = \mu_k N = \mu_k mg \cos\theta$$
### Core Logic
Substitute f_k$f_k$ into the equation of motion:
mg sintheta - mu_k mg costheta = ma$$mg \sin\theta - \mu_k mg \cos\theta = ma$$
Divide by m$m$:
g sintheta - mu_k g costheta = a$$g \sin\theta - \mu_k g \cos\theta = a$$
### Step 1: Substitute Values
Given a = fracg2$a = \frac{g}{2}$ and theta = 60^circ$\theta = 60^\circ$:
g sin 60^circ - mu_k g cos 60^circ = fracg2$$g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2}$$fracsqrt32 - fracmu_k2 = frac12$$\frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2}$$sqrt3 - mu_k = 1 implies mu_k = sqrt3 - 1$$\sqrt{3} - \mu_k = 1 \implies \mu_k = \sqrt{3} - 1$$
### Pattern Recognition
Sees: Block sliding down with acceleration on an incline.
Shortcut: The acceleration on an incline is a = g(sintheta - mu_k costheta)$a = g(\sin\theta - \mu_k \cos\theta)$. For theta = 60^circ$\theta = 60^\circ$, this is a = gleft(fracsqrt32 - fracmu_k2right)$a = g\left(\frac{\sqrt{3}}{2} - \frac{\mu_k}{2}\right)$. Equating to g/2$g/2$ gives mu_k = sqrt3 - 1$\mu_k = sqrt{3} - 1$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q15jee_main_2025_08_april_eveningNewton's Second Law
A body of mass 2mathrm~kg$2\mathrm{~kg}$ moving with velocity of vecv_mathrmin = 3hati +4hatjmathrm~m/s$\vec{v}_{\mathrm{in}} = 3\hat{i} +4\hat{j}\mathrm{~m/s}$ enters into a constant force field of 6mathrm~N$6\mathrm{~N}$ directed along positive z-axis. If the body remains in the field for a period of frac53$\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is:
### Related Formula
vecF = m veca$$\vec{F} = m \vec{a}$$vecv = vecu + vecat$$\vec{v} = \vec{u} + \vec{a}t$$
where,
vecF$\vec{F}$ = constant force vector
m$m$ = mass of body
veca$\vec{a}$ = acceleration vector
vecu$\vec{u}$ = initial velocity vector
vecv$\vec{v}$ = final velocity vector
### Core Logic
Given parameters:
- Mass, m = 2mathrm~kg$m = 2\mathrm{~kg}$
- Initial velocity, vecu = 3hati + 4hatjmathrm~m/s$\vec{u} = 3\hat{i} + 4\hat{j}\mathrm{~m/s}$
- Force, vecF = 6hatkmathrm~N$\vec{F} = 6\hat{k}\mathrm{~N}$ (directed along positive z-axis)
- Time interval, t = frac53mathrm~s$t = \frac{5}{3}\mathrm{~s}$
Calculate the acceleration vector veca$\vec{a}$:
veca = fracvecFm = frac6hatk2 = 3hatkmathrm~m/s^2$$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k}\mathrm{~m/s}^{2}$$
### Step 1: Compute Final Velocity
Using the kinematic equation of motion:
vecv = vecu + vecat$$\vec{v} = \vec{u} + \vec{a}t$$vecv = (3hati + 4hatj) + (3hatk) left(frac53right)$$\vec{v} = (3\hat{i} + 4\hat{j}) + (3\hat{k}) \left(\frac{5}{3}\right)$$vecv = 3hati + 4hatj + 5hatkmathrm~m/s$$\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$$
Thus, the emerging velocity of the body is 3hati + 4hatj + 5hatkmathrm~m/s$3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$.
### Pattern Recognition
Sees: Orthogonal initial velocity and force field direction.
Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged (3hati + 4hatj$3\hat{i} + 4\hat{j}$). Simply compute the z-component change: v_z = a_z t = left(frac62right) left(frac53right) = 5$v_z = a_z t = \left(\frac{6}{2}\right) \left(\frac{5}{3}\right) = 5$. Result: 3hati + 4hatj + 5hatk$3\hat{i} + 4\hat{j} + 5\hat{k}$. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Q12jee_main_2025_29_jan_eveningVariable Mass System
A sand dropper drops sand of massm(t)$m(t)$ on a conveyer belt at a rate proportional to the square root of speed (v$v$) of the belt, i.e. fracdmdt propto sqrtv$\frac{dm}{dt} \propto \sqrt{v}$ . If P$P$ is the power delivered to run the belt at constant speed then which of the following relationship is true?
### Related Formula
F_textthrust = left(fracdmdtright)v$$F_{\text{thrust}} = \left(\frac{dm}{dt}\right)v$$P = F cdot v$P = F \cdot v$
### Core Logic
Given the sand dropping rate rule:
fracdmdt = C sqrtv quad (textwhere C text is a constant)$$\frac{dm}{dt} = C \sqrt{v} \quad (\text{where } C \text{ is a constant})$$
To maintain a constant velocity v$v$, the continuous force applied by the conveyor system must balance the rate of gain of momentum of the dropped sand:
F = left(fracdmdtright) v = (C sqrtv) cdot v = C v^3/2$$F = \left(\frac{dm}{dt}\right) v = (C \sqrt{v}) \cdot v = C v^{3/2}$$
Power delivered is the product of force and speed:
P = F cdot v = (C v^3/2) cdot v = C v^5/2$$P = F \cdot v = (C v^{3/2}) \cdot v = C v^{5/2}$$
Squaring both sides of the expression:
P^2 propto v^5$P^2 \propto v^5$
### Pattern Recognition
In variable mass problems involving dropping dust/sand at rest onto a moving frame, the thrust force always simplifies to v cdot fracdmdt$v \cdot \frac{dm}{dt}$, which makes power scale as v^2 cdot fracdmdt$v^2 \cdot \frac{dm}{dt}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
More Laws of Motion Questions — jee_main_2026_21_jan_morning
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