(A) Given: Mass $m = 1 \,kg$,acceleration $a(x) = \beta x$,where $\beta = 5 \,s^{-2}$.Force $F = m \cdot a = 1 \cdot (5x) = 5x \,N$.Work done $W = \in

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(A) Given: Mass $m = 1 \,kg$,acceleration $a(x) = \beta x$,where $\beta = 5 \,s^{-2}$.Force $F = m \cdot a = 1 \cdot (5x) = 5x \,N$.Work done $W = \int_{x_1}^{x_2} F dx$.Converting units to meters: $x_1 = 2 \,cm = 0.02 \,m$ and $x_2 = 5 \,cm = 0.05 \,m$.$W = \int_{0.02}^{0.05} 5x dx = 5 \left[ \frac{x^2}{2} \right]_{0.02}^{0.05}$.$W = \frac{5}{2} [ (0.05)^2 - (0.02)^2 ]$.$W = 2.5 [ 25 \times 10^{-4} - 4 \times 10^{-4} ]$.$W = 2.5 \times 21 \times 10^{-4} = 52.5 \times 10^{-4} \,J$.

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

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$A$ $1 \,kg$ box placed at the origin starts sliding along the $x$-axis under the action of a force $\vec{F} = F \hat{i}$. Its acceleration as a function of $x$ is given by $a(x) = \beta x$,where $\beta = 5 \,s^{-2}$. The work done by $\vec{F}$ in moving the box from $x = 2 \,cm$ to $x = 5 \,cm$ in joules is:

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A
$52.5 \times 10^{-4}$
B
$105.5 \times 10^{-4}$
C
$17.0 \times 10^{-4}$
D
$34.0 \times 10^{-4}$

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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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$A$ block of mass $2 \ kg$ is initially at rest on a horizontal frictionless surface. $A$ horizontal force $\overrightarrow{F} = (9 - x^2) \hat{i} \ N$ acts on it, starting from $x = 0$. The maximum kinetic energy of the block between $x = 0$ and $x = 3 \ m$ in joules is:

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What is a variable force? Write the equation for the work done by a variable force in one dimension.

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$A$ force acts on a $3.0 \ g$ 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 metres and $t$ is in seconds. The work done during the first $4 \ s$ is ................. $mJ$

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The work done in displacing a particle from $y=a$ to $y=2a$ by a force $F = -\frac{K}{y^2}$ acting along the $y$-axis is:

MediumTS EAMCET 2025

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The graph between power $(P)$ and time $(t)$ is given below. Which of the following options is correct for the work done by the force [from $t = 0$ to time $t$]?

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$A$ block of mass $2 \ kg$ is initially at rest on a horizontal frictionless surface. $A$ horizontal force $\overrightarrow{F} = (9 - x^2) \hat{i} \ N$ acts on it, starting from $x = 0$. The maximum kinetic energy of the block between $x = 0$ and $x = 3 \ m$ in joules is:

What is a variable force? Write the equation for the work done by a variable force in one dimension.

$A$ force acts on a $3.0 \ g$ 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 metres and $t$ is in seconds. The work done during the first $4 \ s$ is ................. $mJ$

The work done in displacing a particle from $y=a$ to $y=2a$ by a force $F = -\frac{K}{y^2}$ acting along the $y$-axis is:

The graph between power $(P)$ and time $(t)$ is given below. Which of the following options is correct for the work done by the force [from $t = 0$ to time $t$]?

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$A$ force acts on a $3.0 \ g$ 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 metres and $t$ is in seconds. The work done during the first $4 \ s$ is ................. $mJ$