(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

Medium

$A$ 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

The kinetic energy acquired by a body of mass $m$ traveling a distance $s$,starting from rest under the action of a constant force,is directly proportional to:

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In the figure shown,all the surfaces are frictionless,and the mass of the block $m$ is $1\,kg$. The block and wedge are held initially at rest. Now,the wedge is given a horizontal acceleration of $5\,m/s^2$ by applying a force on the wedge,such that the block does not slip on the wedge. Then,the work done by the normal force in the ground frame on the block in $2\,sec$ is ............... $J$.

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$A$ boy holds a uniform chain of length $2 \, m$ which is kept on a smooth table such that a length of $60 \, cm$ hangs freely from the edge of the table. The total mass of the chain is $4 \, kg$. What is the work done in pulling the entire chain on the table (in $, J$)? (Take $g = 10 \, m/s^2$)

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$A$ force of $(3\,\hat i + 4\,\hat j) \, N$ acts on a body and displaces it by $(3\,\hat i + 4\,\hat j) \, m$. The work done by the force is .............. $J$.

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$A$ chain of mass $M$ and length $L$ is hanging from the edge of a table such that one-third of its length is hanging off the edge. The work done to pull the hanging part onto the table is:

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The kinetic energy acquired by a body of mass $m$ traveling a distance $s$,starting from rest under the action of a constant force,is directly proportional to:

$A$ boy holds a uniform chain of length $2 \, m$ which is kept on a smooth table such that a length of $60 \, cm$ hangs freely from the edge of the table. The total mass of the chain is $4 \, kg$. What is the work done in pulling the entire chain on the table (in $, J$)? (Take $g = 10 \, m/s^2$)

$A$ force of $(3\,\hat i + 4\,\hat j) \, N$ acts on a body and displaces it by $(3\,\hat i + 4\,\hat j) \, m$. The work done by the force is .............. $J$.

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