(A) The work done by a constant force is given by $W = \vec{F} \cdot \vec{S}$. Since the force acts along the direction of motion,$W = F \cdot S$.Give

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(A) The work done by a constant force is given by $W = \vec{F} \cdot \vec{S}$. Since the force acts along the direction of motion,$W = F \cdot S$.Given the total work $W_{total} = 300 \ J$ and force $F = 50 \ N$,the total distance $S_{total}$ is $S_{total} = \frac{W_{total}}{F} = \frac{300}{50} = 6 \ m$.The motion occurs in three equal spells of distance $d$. Thus,$3d = 6 \ m$,which means $d = 2 \ m$.The particle moves $2 \ m$ along the $x$-axis,$2 \ m$ parallel to the $y$-axis,and $2 \ m$ parallel to the $z$-axis.Therefore,the final coordinates are $(2, 2, 2) \ m$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Constant Force

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$A$ particle is made to move from the origin in three spells of equal distances,first along the $x$-axis,second parallel to the $y$-axis,and third parallel to the $z$-axis. One of the forces acting on it has a constant magnitude of $50 \ N$ and always acts along the direction of motion. Work done by this force in the three spells of motion are equal,and the total work done in all three spells is $300 \ J$. The final coordinates of the particle will be:

A
$(2, 2, 2) \ m$
B
$(4, 4, 4) \ m$
C
$(6, 6, 6) \ m$
D
$(10, 10, 10) \ m$

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Work, Energy, Power and Collision

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Work Done by Constant Force

$A$ body of mass $2 \ kg$ is moving along the $x$-direction such that its displacement as a function of time is given by $x(t) = \alpha t^2 + \beta t + \gamma \ m$, where $\alpha = 1 \ m/s^2$, $\beta = 1 \ m/s$, and $\gamma = 1 \ m$. The work done on the body during the time interval $t = 2 \ s$ to $t = 3 \ s$ is . . . . . . $J$.

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$A$ body of mass $m$ starts from rest and acquires a velocity $v$ in time $t_1$. The work done in time $t$ is proportional to:

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If a force of constant magnitude acts in a direction perpendicular to the motion of a particle,then its

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$A$ particle moves from position $\vec{r}_1 = 3\hat{i} + 2\hat{j} - 6\hat{k}$ to position $\vec{r}_2 = 14\hat{i} + 13\hat{j} + 9\hat{k}$ under the action of force $\vec{F} = 4\hat{i} + \hat{j} + 3\hat{k} \text{ N}$. The work done will be ............ $\text{J}$.

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$A$ particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec{F} = (3\hat{i} + 4\hat{j})$ through two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

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