NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355826
A force acts on a body and displaces it in its direction. The graph shows the relation between the force and displacement, the work done by the force is
1 \(240\;J\)
2 \(360\;J\)
3 \(840\;J\)
4 \(720\;J\)
Explanation:
The work done = Area under \(F-s\) graph. \( \Rightarrow W = \frac{1}{2}(12) \times 60 = 360\;J\)
PHXI06:WORK ENERGY AND POWER
355827
An object of mass \(m\) is tied to a string of length \(L\) and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle \(\theta\) with the vertical. Work done by the force \(F\) is
1 \(m g L(1-\sin \theta)\)
2 \(m g L\)
3 \(m g L(1-\cos \theta)\)
4 \(m g L(1+\cos \theta)\)
Explanation:
The forces that act on the mass are \(F\), \(mg\) and tension (\(T\)). From work energy theorem \(\begin{aligned}& \sum W=\Delta K \\& W_{T}+W_{F}+W_{g}=0\end{aligned}\) \(W_{T}=0\) since tension at a point is always perpendicular to the small displacement of the mass. \(\Rightarrow W_{F}=-W_{g}=\Delta U=m g L(1-\cos \theta)\)
PHXI06:WORK ENERGY AND POWER
355828
A force \(F\) acting on a particle varies with the position \(x\) as shown in the graph. Find the work done by the force in displacing the particle from \(x = - a\,\,{\rm{to}}\,\,x = + 2a\)
1 \(\dfrac{3 a b}{2}\)
2 \(\dfrac{4 a b}{2}\)
3 \(\dfrac{2}{3 a b}\)
4 \(\dfrac{2}{4 a b}\)
Explanation:
The area under force-displacement area gives work done. \(\Rightarrow\) The given interval \(x=-a\) to \(x=+2 a\) can be divided into \(x=-a\) to 0 and \(x=0\) to \(2 b\). \(\Rightarrow W_{1}=-\dfrac{1}{2} a b\) and \(W_{2}=\dfrac{1}{2}(2 a)(2 b)=2 a b\) \(\Rightarrow\) Total work done \(=2 a b-\dfrac{1}{2} a b=\dfrac{3 a b}{2}\).
PHXI06:WORK ENERGY AND POWER
355829
A force \(\vec{F}=[y \hat{i}+x \hat{j}]\) act on a particle moving in \(x\) - \(y\) plane starting from the point \((4,6)\), the particle is taken along straight line to \((7,8)\). The work done by the force is:
1 Zero
2 40
3 32
4 18
Explanation:
\(W=\int \vec{F} \cdot d \vec{S}=\int[(y \hat{i}+x \hat{j}) \cdot(d x \hat{i}+d y \hat{j})]\) \(\begin{aligned}W=\int(y d x+x d y) & =\int_{(4,6)}^{(7,8)} d(x y) \\& =[56-24]=32\end{aligned}\)
355826
A force acts on a body and displaces it in its direction. The graph shows the relation between the force and displacement, the work done by the force is
1 \(240\;J\)
2 \(360\;J\)
3 \(840\;J\)
4 \(720\;J\)
Explanation:
The work done = Area under \(F-s\) graph. \( \Rightarrow W = \frac{1}{2}(12) \times 60 = 360\;J\)
PHXI06:WORK ENERGY AND POWER
355827
An object of mass \(m\) is tied to a string of length \(L\) and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle \(\theta\) with the vertical. Work done by the force \(F\) is
1 \(m g L(1-\sin \theta)\)
2 \(m g L\)
3 \(m g L(1-\cos \theta)\)
4 \(m g L(1+\cos \theta)\)
Explanation:
The forces that act on the mass are \(F\), \(mg\) and tension (\(T\)). From work energy theorem \(\begin{aligned}& \sum W=\Delta K \\& W_{T}+W_{F}+W_{g}=0\end{aligned}\) \(W_{T}=0\) since tension at a point is always perpendicular to the small displacement of the mass. \(\Rightarrow W_{F}=-W_{g}=\Delta U=m g L(1-\cos \theta)\)
PHXI06:WORK ENERGY AND POWER
355828
A force \(F\) acting on a particle varies with the position \(x\) as shown in the graph. Find the work done by the force in displacing the particle from \(x = - a\,\,{\rm{to}}\,\,x = + 2a\)
1 \(\dfrac{3 a b}{2}\)
2 \(\dfrac{4 a b}{2}\)
3 \(\dfrac{2}{3 a b}\)
4 \(\dfrac{2}{4 a b}\)
Explanation:
The area under force-displacement area gives work done. \(\Rightarrow\) The given interval \(x=-a\) to \(x=+2 a\) can be divided into \(x=-a\) to 0 and \(x=0\) to \(2 b\). \(\Rightarrow W_{1}=-\dfrac{1}{2} a b\) and \(W_{2}=\dfrac{1}{2}(2 a)(2 b)=2 a b\) \(\Rightarrow\) Total work done \(=2 a b-\dfrac{1}{2} a b=\dfrac{3 a b}{2}\).
PHXI06:WORK ENERGY AND POWER
355829
A force \(\vec{F}=[y \hat{i}+x \hat{j}]\) act on a particle moving in \(x\) - \(y\) plane starting from the point \((4,6)\), the particle is taken along straight line to \((7,8)\). The work done by the force is:
1 Zero
2 40
3 32
4 18
Explanation:
\(W=\int \vec{F} \cdot d \vec{S}=\int[(y \hat{i}+x \hat{j}) \cdot(d x \hat{i}+d y \hat{j})]\) \(\begin{aligned}W=\int(y d x+x d y) & =\int_{(4,6)}^{(7,8)} d(x y) \\& =[56-24]=32\end{aligned}\)
355826
A force acts on a body and displaces it in its direction. The graph shows the relation between the force and displacement, the work done by the force is
1 \(240\;J\)
2 \(360\;J\)
3 \(840\;J\)
4 \(720\;J\)
Explanation:
The work done = Area under \(F-s\) graph. \( \Rightarrow W = \frac{1}{2}(12) \times 60 = 360\;J\)
PHXI06:WORK ENERGY AND POWER
355827
An object of mass \(m\) is tied to a string of length \(L\) and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle \(\theta\) with the vertical. Work done by the force \(F\) is
1 \(m g L(1-\sin \theta)\)
2 \(m g L\)
3 \(m g L(1-\cos \theta)\)
4 \(m g L(1+\cos \theta)\)
Explanation:
The forces that act on the mass are \(F\), \(mg\) and tension (\(T\)). From work energy theorem \(\begin{aligned}& \sum W=\Delta K \\& W_{T}+W_{F}+W_{g}=0\end{aligned}\) \(W_{T}=0\) since tension at a point is always perpendicular to the small displacement of the mass. \(\Rightarrow W_{F}=-W_{g}=\Delta U=m g L(1-\cos \theta)\)
PHXI06:WORK ENERGY AND POWER
355828
A force \(F\) acting on a particle varies with the position \(x\) as shown in the graph. Find the work done by the force in displacing the particle from \(x = - a\,\,{\rm{to}}\,\,x = + 2a\)
1 \(\dfrac{3 a b}{2}\)
2 \(\dfrac{4 a b}{2}\)
3 \(\dfrac{2}{3 a b}\)
4 \(\dfrac{2}{4 a b}\)
Explanation:
The area under force-displacement area gives work done. \(\Rightarrow\) The given interval \(x=-a\) to \(x=+2 a\) can be divided into \(x=-a\) to 0 and \(x=0\) to \(2 b\). \(\Rightarrow W_{1}=-\dfrac{1}{2} a b\) and \(W_{2}=\dfrac{1}{2}(2 a)(2 b)=2 a b\) \(\Rightarrow\) Total work done \(=2 a b-\dfrac{1}{2} a b=\dfrac{3 a b}{2}\).
PHXI06:WORK ENERGY AND POWER
355829
A force \(\vec{F}=[y \hat{i}+x \hat{j}]\) act on a particle moving in \(x\) - \(y\) plane starting from the point \((4,6)\), the particle is taken along straight line to \((7,8)\). The work done by the force is:
1 Zero
2 40
3 32
4 18
Explanation:
\(W=\int \vec{F} \cdot d \vec{S}=\int[(y \hat{i}+x \hat{j}) \cdot(d x \hat{i}+d y \hat{j})]\) \(\begin{aligned}W=\int(y d x+x d y) & =\int_{(4,6)}^{(7,8)} d(x y) \\& =[56-24]=32\end{aligned}\)
355826
A force acts on a body and displaces it in its direction. The graph shows the relation between the force and displacement, the work done by the force is
1 \(240\;J\)
2 \(360\;J\)
3 \(840\;J\)
4 \(720\;J\)
Explanation:
The work done = Area under \(F-s\) graph. \( \Rightarrow W = \frac{1}{2}(12) \times 60 = 360\;J\)
PHXI06:WORK ENERGY AND POWER
355827
An object of mass \(m\) is tied to a string of length \(L\) and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle \(\theta\) with the vertical. Work done by the force \(F\) is
1 \(m g L(1-\sin \theta)\)
2 \(m g L\)
3 \(m g L(1-\cos \theta)\)
4 \(m g L(1+\cos \theta)\)
Explanation:
The forces that act on the mass are \(F\), \(mg\) and tension (\(T\)). From work energy theorem \(\begin{aligned}& \sum W=\Delta K \\& W_{T}+W_{F}+W_{g}=0\end{aligned}\) \(W_{T}=0\) since tension at a point is always perpendicular to the small displacement of the mass. \(\Rightarrow W_{F}=-W_{g}=\Delta U=m g L(1-\cos \theta)\)
PHXI06:WORK ENERGY AND POWER
355828
A force \(F\) acting on a particle varies with the position \(x\) as shown in the graph. Find the work done by the force in displacing the particle from \(x = - a\,\,{\rm{to}}\,\,x = + 2a\)
1 \(\dfrac{3 a b}{2}\)
2 \(\dfrac{4 a b}{2}\)
3 \(\dfrac{2}{3 a b}\)
4 \(\dfrac{2}{4 a b}\)
Explanation:
The area under force-displacement area gives work done. \(\Rightarrow\) The given interval \(x=-a\) to \(x=+2 a\) can be divided into \(x=-a\) to 0 and \(x=0\) to \(2 b\). \(\Rightarrow W_{1}=-\dfrac{1}{2} a b\) and \(W_{2}=\dfrac{1}{2}(2 a)(2 b)=2 a b\) \(\Rightarrow\) Total work done \(=2 a b-\dfrac{1}{2} a b=\dfrac{3 a b}{2}\).
PHXI06:WORK ENERGY AND POWER
355829
A force \(\vec{F}=[y \hat{i}+x \hat{j}]\) act on a particle moving in \(x\) - \(y\) plane starting from the point \((4,6)\), the particle is taken along straight line to \((7,8)\). The work done by the force is:
1 Zero
2 40
3 32
4 18
Explanation:
\(W=\int \vec{F} \cdot d \vec{S}=\int[(y \hat{i}+x \hat{j}) \cdot(d x \hat{i}+d y \hat{j})]\) \(\begin{aligned}W=\int(y d x+x d y) & =\int_{(4,6)}^{(7,8)} d(x y) \\& =[56-24]=32\end{aligned}\)
