Q61JEE Main 2023NAT

A block of mass $10 {\text{kg}}$ is moving along ${\text{x}}$ -axis under the action of force ${\text{F}}=5 {\text{xN}}$ . The work done by the force in moving the block from ${\text{x}}=2 {\text{m}}$ to $4 {\text{m}}$ will be _______ J. [15-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Work done, ${\text{w}}=∫ {\text{Fdx}}$ ${\text{w}}\;=∫_{2}^{4} 5 {\text{xdx}}$ ${\text{w}}\;=\;\frac{5}{2} \$ [ {\text{x}}^2]$ _2^4{\text{w}};=;\frac{5}{2}(16-4){\text{w}};=30 {\text{J}}$

Q62JEE Main 2023NAT

A car accelerates from rest to $u {\text{m}} / {\ s}$ . The energy spent in this process is ${\text{E}} {\text{J}}$ . The energy required to accelerate the car from ${\text{u}} {\text{m}} / {\ s}$ to $2 {\text{u}} {\text{m}} / {\ s}$ is ${\text{nE}} {\text{J}}$ . The value of ${\text{n}}$ is _______. [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{E}}_1=\;\frac{1}{2} {\mu}^{2-}0=\;\frac{1}{2} {\mu}^2= {\text{E}}$ $\; {\text{E}}_2=\;\frac{1}{2} {\text{m}}(24)^{2-}\;\frac{1}{2} {\mu}^2$ $\;=\;\frac{3}{2} {\mu}^2=3 {\text{E}}$

Q63JEE Main 2022MCQ

A bullet of mass $200 {\text{g}}$ having initial kinetic energy $90 {\text{J}}$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $40 {\text{J}}$ within $1 {\ s}$ , the minimum length of the pool, the bullet has to travel so that it completely comes to rest is [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{1}{2} m x^2=90$ $\Rightarrow \;\frac{1}{2} \times 0.2 \times x^2=90$ $x^2=900$ $x=30 {\text{m}} / {\ s}$ $\;\frac{1}{2} m v^2=40 \Rightarrow v=\;\frac{2}{3} \times 30=20 {\text{m}} / {\ s}$ $20=30-a \times 1 \Rightarrow a=-10 {\text{m}} / {\ s}^2$ $0-x^2=2 a s$ $s=\;\frac{x^2}{-2 a}=\;\frac{30 \times 30}{2 \times 10}$ $=45 {\text{m}}$

Q64JEE Main 2022MCQ

In the given figure, the block of mass $m$ is dropped from the point ' $A$ '. The expression for kinetic energy of block when it reaches point ' $B$ ' is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Loss in potential energy = gain in kinetic energy $- (m g (y-y_0) -m g y) =K E-0$ $\Rightarrow {\text{KE}}= {\text{mgy}}_0$

Q65JEE Main 2022MCQ

Arrange the four graphs in descending order of total work done; where $W_1, W_2, W_3$ and $W_4$ are the work done corresponding to figure $a, b, c$ and $d$ respectively. [26-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{W}}_{ {\text{a}}}=0, {\text{W}}_{ {\text{b}}}=+ {\text{ve}}, {\text{W}}_{ {\text{c}}}=+ {\text{ve}}> {\text{W}}_{ {\text{b}}}, {\text{W}}_{ {\text{d}}}=- {\text{ve}}$ $\Rightarrow {\text{W}}_{ {\text{c}}}> {\text{W}}_{ {\text{b}}}> {\text{W}}_{ {\text{a}}}> {\text{W}}_{ {\text{d}}}$ $\Rightarrow {\text{W}}_3> {\text{W}}_2> {\text{W}}_1> {\text{W}}_4$

Q66JEE Main 2022MCQ

As per the given figure, two blocks each of mass $250 {\text{g}}$ are connected to a spring of spring constant $2 {\text{Nm}}^{-1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is: [26-Jul-2022-Shift-1]

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📖 Explanation

$\because$ Loss in $K E=$ Gain in spring energy $\Rightarrow \;\frac{1}{2} m v^2 \times 2=\;\frac{1}{2} k x_m^2$ $\Rightarrow 2 \times \;\frac{1}{4} \times v^2=2 \times x_m^2$ $\Rightarrow x_m=\sqrt{\;\frac{v^2}{4}}=\;\frac{v}{2}$

Q67JEE Main 2022MCQ

A particle experiences a variable force $\vec{F}=(4 x \hat{i}+3 y^2 \hat{j})$ in a horizontal $x-y$ plane. Assume distance in meters and force is newton. If the particle moves from point $(1,2)$ to point $(2,3)$ in the $x-y$ plane, then Kinetic Energy changes by : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$W=∫ \vec{F} \cdot \vec{d r}$ $=∫_1^2 4 x d x+∫_2^3 3 y^2 d y$ $=\$ [2 x^2]_1^{2+}$[y^3]_2^3=2 \times 3+(27-8)=25 {\text{J}}$

Q68JEE Main 2022NAT

A $0.5 {\text{kg}}$ block moving at a speed of $12 {\text{ms}}^{-1}$ compresses a spring through a distance $30 {\text{cm}}$ when its speed is halved. The spring constant of the spring will be ____ ${\text{Nm}}^{-1}$ [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{1}{2} m V^2=\;\frac{1}{2} k x^{2+}\;\frac{1}{2} m(\;\frac{v}{2})^2$ $\Rightarrow \;\frac{3}{8} m v^2=\;\frac{1}{2} k x^2$ $\Rightarrow k=\;\frac{3}{4} \times \;\frac{1}{2} \times \;\frac{144}{9} \times 100$ $=600$ $\Rightarrow 600$

Q69JEE Main 2022MCQ

A bag of sand of mass $9.8 {kg}$ is suspended by a rope. A bullet of $200 {g}$ travelling with speed $10 {ms}^{-1}$ gets embedded in it, then loss of kinetic energy will be : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Loss in $K E=\;\frac{1}{2} \times \;\frac{m_{1} m_{2}}{m_{1}+m_{2}} \times v^{2}$ $=\;\frac{1}{2} \times \;\frac{9.8 \times 0.2}{10} \times (10)^{2}$ $=9.8 {J}$

Q70JEE Main 2022MCQ

A body of mass $0.5 {kg}$ travels on straight line path with velocity $v=(3 x^{2}+4) {m} / {s}$ . The net workdone by the force during its displacement from $x=0$ to $x=2 {m}$ is : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v=3 x^{2}+4$ at $x=0, v_{1}=4 {m} / {s}$ $x=2, v_{n}-16 {m} / {s}$ $\Rightarrow$ Work done $=\Delta$ kinetic energy $-\;\frac{1}{2} \times m(v_{2}^{2}-v_{1}^{2})$ $=\;\frac{1}{4}(256-16)$ $=60 {J}$

Q71JEE Main 2022MCQ

A particle of mass $500 {\text{gm}}$ is moving in a straight line with velocity $v=b x^{5 / 2}$ . The work done by the net force during its displacement from $x=0$ to $x=4 m$ is : (Take $.b=0.25 m^{-3 / 2} s^{-1})$ . [29-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$W_{\;\text{total }\;}=\Delta K$ $=\;\frac{1}{2} (\;\frac{1}{2}) [ \{b(4)^{5 / 2}\} ^{2-}0]$ $=\;\frac{b^2}{4} \times 4^5$ $\Rightarrow W_{\;\text{total }\;}=16 J$

Q72JEE Main 2022MCQ

A ball is projected with kinetic energy $E$ , at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become : [29-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{E}}=\;\frac{1}{2} {\mu}^2$ At Highest point, Velocity ${\text{V}}= {\text{u}} \cos 60^{\circ}=\;{ {\text{u}}}/{2}$ $\therefore$ K.E at topmost point $=\;\frac{1}{2} {\text{m}} (\;{ {\text{u}}}/{2}) ^2=\;{ {\text{E}}}/{4}$

Q73JEE Main 2022MCQ

Sand is being dropped from a stationary dropper at a rate of $0.5 {\text{kgs}}^{-1}$ on a conveyor belt moving with a velocity of $5 {\text{ms}}^{-1}$ . The power needed to keep the belt moving with the same velocity will be : [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d m}{d t}=0.5 {\text{kg}} / {\text{s}}$ $v=5 {\text{m}} / {\text{s}}$ $F=\;\frac{v d m}{d t}=2.5 {\text{kg}} {\text{m}} / {\text{s}}^2$ $P=\overline{F} \cdot \overline{v}=(2.5)(5) {\text{W}}$ $=12.5 {\text{W}}$

Q74JEE Main 2022NAT

A ball of mass $100 {\text{g}}$ is dropped from a height $h=10 {\text{cm}}$ on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\;\frac{h}{2}$ . The spring constant is ${\text{Nm}}^{-1}$ . (Use ${\text{g}}=10 {\text{ms}}^{-2}$ ) [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$m g(h+\;\frac{h}{2})=\;\frac{1}{2} k(\;\frac{h}{2})^2$ $\Rightarrow 0.1 \times 10 \times (0.15)=\;\frac{1}{2} k(0.05)^2$ $\Rightarrow k=120 {\text{N}} / {\text{m}}$

Q75JEE Main 2022NAT

A block of mass 'Im' (as shown in figure) moving with kinetic energy E compresses a spring through a distance $25 {\text{cm}}$ when, its speed is halved. The value of spring constant of used spring will be ${\text{nE}} {\text{Nm}}{ }^{-1}$ for ${\text{n}}=$ _______. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Using work - energy theorem ${\text{W}}_{ {\text{net}}}=( {\text{K}}_{ {\text{f}}}- {\text{K}}_{ {\text{i}}})$ $\Rightarrow -\;\frac{1}{2} {\text{Kx}}^2=\;\frac{1}{2} {\text{m}}(\;{ {\text{v}}}/{2})^{2-}\;\frac{1}{2} {\ mv}^2=\;{ {\text{E}}}/{4}- {\text{E}}$ $\Rightarrow \;\frac{1}{2} {\text{Kx}}^2=\;{3 {\text{E}}}/{4} \Rightarrow {\text{K}}=\;{3 {\text{E}}}/{2 {\text{x}}^2}$ $\Rightarrow {\text{K}}=\;{3 {\text{E}}}/{2 \times (\;\frac{1}{4})^2}=24 {\text{E}}$ ${\text{n}}=24$

Q76JEE Main 2022NAT

A uniform chain of $6 {\text{m}}$ length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is $0.5$ , the maximum length of the chain hanging from the table is_____ m. [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$(x) g \lambda =\mu (6-x) g \lambda$ where ${\text{x}}$ is length of hanging part $\Rightarrow x=3-0.5 x$ $\Rightarrow x=2 {\text{m}}$

Q77JEE Main 2021NAT

A uniform chain of length $3 {\text{m}}$ and mass $3 {\text{kg}}$ overhangs a smooth table with $2 {\text{m}}$ laying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is ......... . (Take, $g=10 {\text{m}}/ {\text{s}}^{2}$ )

🚀 Solve in Practice Mode

📖 Explanation

The given situation is shown below as Applying law of conservation of energy, we get $\text{K}_{{\text{initial} }}+\text{U}_{{\text{initial} }}=\text{K}_{{\text{final} }}+\text{U}_{{\text{final} }}$ $\text{K}_{{\text{initial} }}=0$ $\Rightarrow \text{U}_{{\text{initial} }}=-1 \times 10 \times \frac{1}{2}$ ( $1 {\text{m}}$ lie below DATUM level, so negative) $\text{K}_{{\text{final} }}=?$ $\text{U}_{{\text{final} }}=-3 \times 10 \times \frac{3}{2}(3 {\text{m}}$ lie below DATUMlevel, so negative) $0-1 \times 10 \times \frac{1}{2}=\text{K}_{ {\text{final} }}-3 \times 10 \times \frac{3}{2}$ $\Rightarrow -5=\text{K}_{{\text{final} }}-45$ $\Rightarrow \text{K}_{{\text{final} }}=45-5$ = 40 J

Q78JEE Main 2021MCQ

Given below is the plot of a potential energy function $U(x)$ for a system, in which a particle is in one dimensional motion, while a conservative force ${F}({x})$ acts on it. Suppose that ${E}_{\text{mech }}=8 {J}$ , the incorrect statement for this system is:

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📖 Explanation

${E}_{\text{mech }}=8 {J}$ (A) at ${x}>{x}_{4}, \;\;\;{U}=$ constant $=6 {J}$ ${K}={E}_{\text{mech. }}-{U}=2 {J}=$ constant (B) at ${x} < {x}_{{1}}, \;\;\;{U}=$ constant $=8 {J}$ ${K}={E}_{\text{mech. }}-{U}=8-8=0 {J}$ Particle is at rest. (C) At ${x}={x}_{2},\;\;\; {U}=0 \$ $\Rightarrow {E}_{\text{mech. }}={K}=8 {J}$ KE is greatest, and particle is moving at fastest speed. (D) At ${x}={x}_{3}, \;\;\;{U}=4 {J}$ ${U}+{K}=8 {J}$ ${K}=4 {J}$

Q79JEE Main 2021MCQ

A porter lifts a heavy suitcase of mass $80 {kg}$ and at the destination lowers it down by a distance of $80 {cm}$ with a constant velocity. Calculate the workdone by the porter in lowering the suitcase. $($ take $g=9.8 {ms}^{-2})$

🚀 Solve in Practice Mode

📖 Explanation

${W}_{\text{Porter }}+{W}_{{mg}}=\Delta{K} . {E} .=0$ ${W}_{\text{Porter }}=-{W}_{{mg}}=-{mgh}$ $=-80 \\times 9.8 \\times .8=-627.2 { J}$

Q80JEE Main 2021NAT

An engine is attached to a wagon through a shock absorber of length $1.5 {\text{m}}$ . The system with a total mass of $40000 {\text{kg}}$ is moving with a speed of $72 {\text{kmh}}^{-1}$ , when the brakes are applied to bring it to rest. In the process of the system beingbrought to rest, the spring of the shock absorber gets compressed by $1.0 {\text{m}}$ . If $90 \%$ of energy of the wagon is lost due to friction, the spring constant is ............. × $10^{5} {\text{N}} / {\text{m}}$ .

🚀 Solve in Practice Mode

📖 Explanation

Given, the length of the shock absorber, I = 1.5 m The total mass of the system, M = 40000 kg The speed of the wagon, v = 72 km/h When brakes are applied, the final velocity, $\text{v}_\text{f}$ = 0 The compressed spring of the shock absorber, x = 1mApplying the work-energy theorem, Work done by the system = Change in kinetic energy W = Δ KE $\text{W}_{{\text{friction} }}+\text{W}_{{\text{spring} }}=\frac{1}{2} \text{m} \text{v}_{\text{f}}^{2}+\frac{1}{2} \text{m} \text{v}_{\text{i}}^{2}$ $-\frac{90}{100}(\frac{1}{2} \text{m} \text{v}^{2})+\text{W}_{{\text{spring} }}=0-\frac{1}{2} \text{m} \text{v}_{\text{i}}^{2}$ (\because 90% energy lost due to friction) $\text{W}_{{\text{spring} }}=-\frac{10}{100} \times \frac{1}{2} \text{m} \text{v}^{2}$ $-\frac{1}{2} \text{k} \text{x}^{2}=\frac{1}{20} \text{m} \text{v}^{2}$ $\text{k}=\frac{\text{m} \text{v}^{2}}{10 \times x^{2}}$ Substituting the values in the above equation, we get $\text{k}=\frac{40000 \times (72 \times \frac{5}{18})^{2}}{10(1)^{2}}$ = 16 × $10^5$ N / m Comparing the spring constant, k = x × $10^5$ The value of the x = 16.