02. Conservation of Energy and Work Energy Theorem
Work, Energy and Power

149120 Some of the following equations are kinetic equations, where the symbols have their usual meaning. The work-energy theorem is represented by

1 $\mathrm{v}=\mathrm{u}+$ at
2 $s=u t$
3 $\mathrm{s}=u t+\frac{1}{2} a t^{2}$
4 $v^{2}=\frac{u^{2}}{2}+a s$
5 $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
Work, Energy and Power

149122 A bead of mass $m$ can slide without friction on a fixed circular horizontal ring of radius $3 R$ having centre at the point $C$. The bead is attached to one of the ends of spring of spring constant $k$. Natural length of spring is $R$ and the other end of the spring is fixed at point $O$ as shown in figure. Bead is released from position $A$, what will be kinetic energy of the bead when it reaches at point $B$ ?

1 $\frac{25}{2} \mathrm{kR}^{2}$
2 $\frac{9}{2} \mathrm{kR}^{2}$
3 $8 \mathrm{kR}^{2}$
4 $12 \mathrm{kR}^{2}$
Work, Energy and Power

149124 A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x=0$ has kinetic energy of $25 \mathrm{~J}$, then the kinetic energy of the particle at $x=16 \mathrm{~m}$ is

1 $45 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $70 \mathrm{~J}$
4 $135 \mathrm{~J}$
5 $20 \mathrm{~J}$
Work, Energy and Power

149125 A $3 \mathrm{~kg}$ object has initial velocity $(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. The total work done on the object if its velocity changes to $(8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ is

1 $216 \mathrm{~J}$
2 $44 \mathrm{~J}$
3 $60 \mathrm{~J}$
4 $120 \mathrm{~J}$
Work, Energy and Power

149120 Some of the following equations are kinetic equations, where the symbols have their usual meaning. The work-energy theorem is represented by

1 $\mathrm{v}=\mathrm{u}+$ at
2 $s=u t$
3 $\mathrm{s}=u t+\frac{1}{2} a t^{2}$
4 $v^{2}=\frac{u^{2}}{2}+a s$
5 $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
Work, Energy and Power

149122 A bead of mass $m$ can slide without friction on a fixed circular horizontal ring of radius $3 R$ having centre at the point $C$. The bead is attached to one of the ends of spring of spring constant $k$. Natural length of spring is $R$ and the other end of the spring is fixed at point $O$ as shown in figure. Bead is released from position $A$, what will be kinetic energy of the bead when it reaches at point $B$ ?

1 $\frac{25}{2} \mathrm{kR}^{2}$
2 $\frac{9}{2} \mathrm{kR}^{2}$
3 $8 \mathrm{kR}^{2}$
4 $12 \mathrm{kR}^{2}$
Work, Energy and Power

149124 A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x=0$ has kinetic energy of $25 \mathrm{~J}$, then the kinetic energy of the particle at $x=16 \mathrm{~m}$ is

1 $45 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $70 \mathrm{~J}$
4 $135 \mathrm{~J}$
5 $20 \mathrm{~J}$
Work, Energy and Power

149125 A $3 \mathrm{~kg}$ object has initial velocity $(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. The total work done on the object if its velocity changes to $(8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ is

1 $216 \mathrm{~J}$
2 $44 \mathrm{~J}$
3 $60 \mathrm{~J}$
4 $120 \mathrm{~J}$
Work, Energy and Power

149120 Some of the following equations are kinetic equations, where the symbols have their usual meaning. The work-energy theorem is represented by

1 $\mathrm{v}=\mathrm{u}+$ at
2 $s=u t$
3 $\mathrm{s}=u t+\frac{1}{2} a t^{2}$
4 $v^{2}=\frac{u^{2}}{2}+a s$
5 $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
Work, Energy and Power

149122 A bead of mass $m$ can slide without friction on a fixed circular horizontal ring of radius $3 R$ having centre at the point $C$. The bead is attached to one of the ends of spring of spring constant $k$. Natural length of spring is $R$ and the other end of the spring is fixed at point $O$ as shown in figure. Bead is released from position $A$, what will be kinetic energy of the bead when it reaches at point $B$ ?

1 $\frac{25}{2} \mathrm{kR}^{2}$
2 $\frac{9}{2} \mathrm{kR}^{2}$
3 $8 \mathrm{kR}^{2}$
4 $12 \mathrm{kR}^{2}$
Work, Energy and Power

149124 A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x=0$ has kinetic energy of $25 \mathrm{~J}$, then the kinetic energy of the particle at $x=16 \mathrm{~m}$ is

1 $45 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $70 \mathrm{~J}$
4 $135 \mathrm{~J}$
5 $20 \mathrm{~J}$
Work, Energy and Power

149125 A $3 \mathrm{~kg}$ object has initial velocity $(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. The total work done on the object if its velocity changes to $(8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ is

1 $216 \mathrm{~J}$
2 $44 \mathrm{~J}$
3 $60 \mathrm{~J}$
4 $120 \mathrm{~J}$
Work, Energy and Power

149120 Some of the following equations are kinetic equations, where the symbols have their usual meaning. The work-energy theorem is represented by

1 $\mathrm{v}=\mathrm{u}+$ at
2 $s=u t$
3 $\mathrm{s}=u t+\frac{1}{2} a t^{2}$
4 $v^{2}=\frac{u^{2}}{2}+a s$
5 $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
Work, Energy and Power

149122 A bead of mass $m$ can slide without friction on a fixed circular horizontal ring of radius $3 R$ having centre at the point $C$. The bead is attached to one of the ends of spring of spring constant $k$. Natural length of spring is $R$ and the other end of the spring is fixed at point $O$ as shown in figure. Bead is released from position $A$, what will be kinetic energy of the bead when it reaches at point $B$ ?

1 $\frac{25}{2} \mathrm{kR}^{2}$
2 $\frac{9}{2} \mathrm{kR}^{2}$
3 $8 \mathrm{kR}^{2}$
4 $12 \mathrm{kR}^{2}$
Work, Energy and Power

149124 A particle is acted upon by a force $F$ which varies with position $x$ as shown in figure. If the particle at $x=0$ has kinetic energy of $25 \mathrm{~J}$, then the kinetic energy of the particle at $x=16 \mathrm{~m}$ is

1 $45 \mathrm{~J}$
2 $30 \mathrm{~J}$
3 $70 \mathrm{~J}$
4 $135 \mathrm{~J}$
5 $20 \mathrm{~J}$
Work, Energy and Power

149125 A $3 \mathrm{~kg}$ object has initial velocity $(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. The total work done on the object if its velocity changes to $(8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$ is

1 $216 \mathrm{~J}$
2 $44 \mathrm{~J}$
3 $60 \mathrm{~J}$
4 $120 \mathrm{~J}$