355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

355809 The figure shows all the surfaces are frictionless and mass of the block,\(m = 1\,kg\). The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of \(10\,m/{s^2}\) by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in \(\sqrt{3}\) seconds is

355810 A force \(\vec{F}=k[y \hat{i}+x \hat{j}]\) where \(k\) is a positive constant acts on a particle moving in \(x\) - \(y\) plane starting from the point \((3,5)\), the particle is taken along a straight line to \((5,7)\). The work done by the force is:

355811 A particle moves under the effect of a force \(F=c x^{2}\) from \(x=0\) to \(x=x_{0}\). The work done in the process is

355812 The displacement \(x\) in metre of a particle of mass \(m\) \(kg\) moving in one dimension under the action of a force is related to the time \(t\) in second by the equation \(t=\sqrt{x}+3\), the work done by the force (in joule) in first six seconds is

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

355809 The figure shows all the surfaces are frictionless and mass of the block,\(m = 1\,kg\). The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of \(10\,m/{s^2}\) by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in \(\sqrt{3}\) seconds is

355810 A force \(\vec{F}=k[y \hat{i}+x \hat{j}]\) where \(k\) is a positive constant acts on a particle moving in \(x\) - \(y\) plane starting from the point \((3,5)\), the particle is taken along a straight line to \((5,7)\). The work done by the force is:

355811 A particle moves under the effect of a force \(F=c x^{2}\) from \(x=0\) to \(x=x_{0}\). The work done in the process is

355812 The displacement \(x\) in metre of a particle of mass \(m\) \(kg\) moving in one dimension under the action of a force is related to the time \(t\) in second by the equation \(t=\sqrt{x}+3\), the work done by the force (in joule) in first six seconds is

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

355809 The figure shows all the surfaces are frictionless and mass of the block,\(m = 1\,kg\). The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of \(10\,m/{s^2}\) by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in \(\sqrt{3}\) seconds is

355810 A force \(\vec{F}=k[y \hat{i}+x \hat{j}]\) where \(k\) is a positive constant acts on a particle moving in \(x\) - \(y\) plane starting from the point \((3,5)\), the particle is taken along a straight line to \((5,7)\). The work done by the force is:

355811 A particle moves under the effect of a force \(F=c x^{2}\) from \(x=0\) to \(x=x_{0}\). The work done in the process is

355812 The displacement \(x\) in metre of a particle of mass \(m\) \(kg\) moving in one dimension under the action of a force is related to the time \(t\) in second by the equation \(t=\sqrt{x}+3\), the work done by the force (in joule) in first six seconds is

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

355809 The figure shows all the surfaces are frictionless and mass of the block,\(m = 1\,kg\). The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of \(10\,m/{s^2}\) by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in \(\sqrt{3}\) seconds is

355810 A force \(\vec{F}=k[y \hat{i}+x \hat{j}]\) where \(k\) is a positive constant acts on a particle moving in \(x\) - \(y\) plane starting from the point \((3,5)\), the particle is taken along a straight line to \((5,7)\). The work done by the force is:

355811 A particle moves under the effect of a force \(F=c x^{2}\) from \(x=0\) to \(x=x_{0}\). The work done in the process is

355812 The displacement \(x\) in metre of a particle of mass \(m\) \(kg\) moving in one dimension under the action of a force is related to the time \(t\) in second by the equation \(t=\sqrt{x}+3\), the work done by the force (in joule) in first six seconds is

355808 A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?

355809 The figure shows all the surfaces are frictionless and mass of the block,\(m = 1\,kg\). The block and wedge are held initially at rest. Now wedge is given a horizontal acceleartion of \(10\,m/{s^2}\) by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in \(\sqrt{3}\) seconds is

355810 A force \(\vec{F}=k[y \hat{i}+x \hat{j}]\) where \(k\) is a positive constant acts on a particle moving in \(x\) - \(y\) plane starting from the point \((3,5)\), the particle is taken along a straight line to \((5,7)\). The work done by the force is:

355811 A particle moves under the effect of a force \(F=c x^{2}\) from \(x=0\) to \(x=x_{0}\). The work done in the process is

355812 The displacement \(x\) in metre of a particle of mass \(m\) \(kg\) moving in one dimension under the action of a force is related to the time \(t\) in second by the equation \(t=\sqrt{x}+3\), the work done by the force (in joule) in first six seconds is