(D) Given: Mass $m = 3 \text{ kg}$, Displacement $x = \frac{t^3}{3} \text{ m}$.First, find the velocity $v$ by differentiating displacement with respe

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) Given: Mass $m = 3 \text{ kg}$, Displacement $x = \frac{t^3}{3} \text{ m}$.First, find the velocity $v$ by differentiating displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \text{ m/s}$.At $t = 0$, $v_i = 0^2 = 0 \text{ m/s}$.At $t = 2 \text{ s}$, $v_f = 2^2 = 4 \text{ m/s}$.According to the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy:$W = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.Substituting the values:$W = \frac{1}{2} \times 3 \times (4)^2 - \frac{1}{2} \times 3 \times (0)^2 = \frac{1}{2} \times 3 \times 16 = 24 \text{ J}$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

Easy

$A$ body of mass $3 \text{ kg}$ is moving under the action of a force which causes a displacement of $\left(\frac{t^3}{3}\right) \text{ m}$, where '$t$' is time in seconds. The work done by the force in the first $2 \text{ seconds}$ is: (in $\text{ J}$)

TS EAMCET 2023 All TS EAMCET

A
$2$
B
$3.8$
C
$5.2$
D
$24$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Work, Energy, Power and Collision

Subtopic

Work Energy Theorem and Conservation of Mechanical Energy

Previous Year

TS EAMCET 2023 Paper All TS EAMCET years →

$A$ bob of a pendulum of length $0.5 \ m$ has a speed of $6 \ ms^{-1}$ at its lowest point. Find the speed of the bob when the string of the pendulum makes $60^{\circ}$ with the vertical. (Take $g=10 \ ms^{-2}$)

MediumAP EAMCET 2021

View Solution

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

EasyTS EAMCET 2020

View Solution

$A$ pendulum bob on a $2\, m$ string is displaced $60^\circ$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path? (Take $g = 9.8\, m/s^2$)

Medium

View Solution

The mass of a pendulum bob is $50 \ gm$. The bob is released from a horizontal position $A$ as shown in the figure. If the length of the pendulum is $1.5 \ m$,what will be its kinetic energy when it reaches the lowest point $B$ of its path? (Take $g = 10 \ m/s^2$)

Easy

View Solution

$A$ force acts on a $3\, g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$,where $x$ is in $meters$ and $t$ is in $seconds$. The work done during the first $4\, seconds$ is ................. $mJ$.

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ body of mass $3 \text{ kg}$ is moving under the action of a force which causes a displacement of $\left(\frac{t^3}{3}\right) \text{ m}$, where '$t$' is time in seconds. The work done by the force in the first $2 \text{ seconds}$ is: (in $\text{ J}$)

Similar Questions

$A$ bob of a pendulum of length $0.5 \ m$ has a speed of $6 \ ms^{-1}$ at its lowest point. Find the speed of the bob when the string of the pendulum makes $60^{\circ}$ with the vertical. (Take $g=10 \ ms^{-2}$)

Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

$A$ pendulum bob on a $2\, m$ string is displaced $60^\circ$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path? (Take $g = 9.8\, m/s^2$)

The mass of a pendulum bob is $50 \ gm$. The bob is released from a horizontal position $A$ as shown in the figure. If the length of the pendulum is $1.5 \ m$,what will be its kinetic energy when it reaches the lowest point $B$ of its path? (Take $g = 10 \ m/s^2$)

$A$ force acts on a $3\, g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$,where $x$ is in $meters$ and $t$ is in $seconds$. The work done during the first $4\, seconds$ is ................. $mJ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module