(D) Given: Mass $m = 3 \,kg$, displacement $s = \frac{t^3}{3}$.Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^3}{3}) = t^2$.Acceleration $a = \fra

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(D) Given: Mass $m = 3 \,kg$, displacement $s = \frac{t^3}{3}$.Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^3}{3}) = t^2$.Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t$.Force $F = ma = 3 \times 2t = 6t$.Work done $W = \int F \cdot ds = \int_0^2 F \cdot v dt = \int_0^2 (6t)(t^2) dt$.$W = \int_0^2 6t^3 dt = 6 \left[ \frac{t^4}{4} \right]_0^2$.$W = 6 \times \frac{16}{4} = 6 \times 4 = 24 \,J$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Variable Force and Force-Displacement Graph

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The displacement $s$ of a body of mass $3 \,kg$ under the action of a force is given by $s = \frac{t^3}{3}$, where $s$ is in metres and $t$ is in seconds. The work done by the force in the first two seconds is (in $\,J$)

TS EAMCET 2023 All TS EAMCET

A
$32$
B
$3.8$
C
$5.2$
D
$24$

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Work, Energy, Power and Collision

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Work Done by Variable Force and Force-Displacement Graph

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The displacement $s$ of a body of mass $3 \,kg$ under the action of a force is given by $s = \frac{t^3}{3}$, where $s$ is in metres and $t$ is in seconds. The work done by the force in the first two seconds is (in $\,J$)

Similar Questions

The figure shows the relationship between force and position. What is the work done by the force for the displacement of the object from $x = 1 \; cm$ to $x = 5 \; cm$ in ergs?

$A$ force is applied to a particle of mass $0.1 \ kg$ as a function of distance as shown in the figure. If it starts from rest at $x = 0$,its velocity at $x = 12 \ m$ is ....... $m/s$.

$A$ force of $(6x^2 - 4x + 3) \text{ N}$ acts on a body of mass $0.75 \text{ kg}$ and displaces it from $x = 2 \text{ m}$ to $x = 5 \text{ m}$. The work done by the force is (in $\text{ J}$)

$A$ graph of potential energy $V(x)$ versus $x$ is shown in the figure. $A$ particle of energy $E_0$ is executing motion in it. Draw the graph of velocity and kinetic energy versus $x$ for one complete cycle $AFA$.

$A$ block of mass $10 \, kg$,moving in $x$ direction with a constant speed of $10 \, m/s$,is subjected to a retarding force $F = 0.1 \, x \, N$ during its travel from $x = 20 \, m$ to $30 \, m$. Its final $KE$ will be ............... $J$.

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