268644
An object has a displacement from position vector \(\vec{r}_{1}=(2 \hat{i}+3 \hat{j}) \mathbf{m}\) to \(\vec{r}_{2}=(4 \hat{i}+6 \hat{j}) \mathbf{m}\) under a force \(\vec{F}=\left(3 x^{2} \hat{i}+2 y \hat{j}\right) \mathbf{N}\), then work done by the force is
1 \(24 \mathrm{~J}\)
2 \(33 \mathrm{~J}\)
3 \(83 \mathrm{~J}\)
4 \(45 \mathrm{~J}\)
Explanation:
\(W=\int d w=\int_{x_{1}}^{x_{2}} F_{x} d x+\int_{y_{1}}^{y_{2}} F_{y} d y\)
Work, Energy and Power
268707
A force \(\mathbf{F}=(2+x) \mathbf{N}\) acts on a particle in \(x\) direction where ' \(x\) ' is in metre. The work done by this force during a displacement from \(x=1 \mathbf{m}\) to \(x=2 \mathbf{m}\) is
1 \(2 \mathrm{~J}\)
2 \(3.5 \mathrm{~J}\)
3 \(4.5 \mathrm{~J}\)
4 \(5 \mathrm{~J}\)
Explanation:
\(W=\int_{1}^{2} F d x\)
Work, Energy and Power
268759
A body is displaced from \((0,0)\) to \((1 \mathrm{~m}, 1 \mathrm{~m})\) along the path \(\mathrm{x}=\mathrm{y}\) by a force \(F=x^{2} \hat{j}+y i \hat{\theta} \hat{n}\). The work done by this force will be
1 \(\frac{4}{3} J\)
2 \(\frac{5}{6} J\)
3 \(\frac{3}{2} J\)
4 \(\frac{7}{5} J\)
Explanation:
\(W=\int_{(0,0)}^{(1,1)} \vec{F} \cdot \overrightarrow{d s} ;\) Here \(\overrightarrow{d s}=d x \hat{i}+d y \hat{j}+d z \hat{k}\)
\(\therefore W=\int_{(0,0)}^{(1,1)}\left(x^{2} d y+y d x\right)=\int_{(0,0)}^{(1,1)}\left(y^{2} d y+x \cdot d x\right)\) (as \(\mathrm{x}=\mathrm{y})\)
Work, Energy and Power
268760
A particle moves under the effect of a force\(F=C x\) from \(x=0\) to \(x=x_{1}\). The work done in the process is (treat \(C\) as a constant)
1 \(\mathrm{Cx}_{1}^{2}\)
2 \(C^{2} / x_{1}^{2}\)
3 \(1 / 2 \mathrm{Cx}_{1}^{2}\)
4 \(1 / 2 C^{2} / x_{1}^{2}\)
Explanation:
\(W=\int_{0}^{X_{1}} F d x\)
Work, Energy and Power
268761
Under the action of a force, a \(2 \mathbf{k g}\) body moves such that its position ' \(x\) ' in meters as a function of time ' \(t\) ' in seconds given by:
\(x=t^{2} / 2\). The work done by the force in the first 5 seconds is
1 \(2.5 \mathrm{~J}\)
2 \(0.25 \mathrm{~J}\)
3 \(25 \mathrm{~J}\)
4 \(250 \mathrm{~J}\)
Explanation:
\(W=\frac{1}{2} m v^{2}\) where \(v=\frac{d x}{d t}\)
268644
An object has a displacement from position vector \(\vec{r}_{1}=(2 \hat{i}+3 \hat{j}) \mathbf{m}\) to \(\vec{r}_{2}=(4 \hat{i}+6 \hat{j}) \mathbf{m}\) under a force \(\vec{F}=\left(3 x^{2} \hat{i}+2 y \hat{j}\right) \mathbf{N}\), then work done by the force is
1 \(24 \mathrm{~J}\)
2 \(33 \mathrm{~J}\)
3 \(83 \mathrm{~J}\)
4 \(45 \mathrm{~J}\)
Explanation:
\(W=\int d w=\int_{x_{1}}^{x_{2}} F_{x} d x+\int_{y_{1}}^{y_{2}} F_{y} d y\)
Work, Energy and Power
268707
A force \(\mathbf{F}=(2+x) \mathbf{N}\) acts on a particle in \(x\) direction where ' \(x\) ' is in metre. The work done by this force during a displacement from \(x=1 \mathbf{m}\) to \(x=2 \mathbf{m}\) is
1 \(2 \mathrm{~J}\)
2 \(3.5 \mathrm{~J}\)
3 \(4.5 \mathrm{~J}\)
4 \(5 \mathrm{~J}\)
Explanation:
\(W=\int_{1}^{2} F d x\)
Work, Energy and Power
268759
A body is displaced from \((0,0)\) to \((1 \mathrm{~m}, 1 \mathrm{~m})\) along the path \(\mathrm{x}=\mathrm{y}\) by a force \(F=x^{2} \hat{j}+y i \hat{\theta} \hat{n}\). The work done by this force will be
1 \(\frac{4}{3} J\)
2 \(\frac{5}{6} J\)
3 \(\frac{3}{2} J\)
4 \(\frac{7}{5} J\)
Explanation:
\(W=\int_{(0,0)}^{(1,1)} \vec{F} \cdot \overrightarrow{d s} ;\) Here \(\overrightarrow{d s}=d x \hat{i}+d y \hat{j}+d z \hat{k}\)
\(\therefore W=\int_{(0,0)}^{(1,1)}\left(x^{2} d y+y d x\right)=\int_{(0,0)}^{(1,1)}\left(y^{2} d y+x \cdot d x\right)\) (as \(\mathrm{x}=\mathrm{y})\)
Work, Energy and Power
268760
A particle moves under the effect of a force\(F=C x\) from \(x=0\) to \(x=x_{1}\). The work done in the process is (treat \(C\) as a constant)
1 \(\mathrm{Cx}_{1}^{2}\)
2 \(C^{2} / x_{1}^{2}\)
3 \(1 / 2 \mathrm{Cx}_{1}^{2}\)
4 \(1 / 2 C^{2} / x_{1}^{2}\)
Explanation:
\(W=\int_{0}^{X_{1}} F d x\)
Work, Energy and Power
268761
Under the action of a force, a \(2 \mathbf{k g}\) body moves such that its position ' \(x\) ' in meters as a function of time ' \(t\) ' in seconds given by:
\(x=t^{2} / 2\). The work done by the force in the first 5 seconds is
1 \(2.5 \mathrm{~J}\)
2 \(0.25 \mathrm{~J}\)
3 \(25 \mathrm{~J}\)
4 \(250 \mathrm{~J}\)
Explanation:
\(W=\frac{1}{2} m v^{2}\) where \(v=\frac{d x}{d t}\)
268644
An object has a displacement from position vector \(\vec{r}_{1}=(2 \hat{i}+3 \hat{j}) \mathbf{m}\) to \(\vec{r}_{2}=(4 \hat{i}+6 \hat{j}) \mathbf{m}\) under a force \(\vec{F}=\left(3 x^{2} \hat{i}+2 y \hat{j}\right) \mathbf{N}\), then work done by the force is
1 \(24 \mathrm{~J}\)
2 \(33 \mathrm{~J}\)
3 \(83 \mathrm{~J}\)
4 \(45 \mathrm{~J}\)
Explanation:
\(W=\int d w=\int_{x_{1}}^{x_{2}} F_{x} d x+\int_{y_{1}}^{y_{2}} F_{y} d y\)
Work, Energy and Power
268707
A force \(\mathbf{F}=(2+x) \mathbf{N}\) acts on a particle in \(x\) direction where ' \(x\) ' is in metre. The work done by this force during a displacement from \(x=1 \mathbf{m}\) to \(x=2 \mathbf{m}\) is
1 \(2 \mathrm{~J}\)
2 \(3.5 \mathrm{~J}\)
3 \(4.5 \mathrm{~J}\)
4 \(5 \mathrm{~J}\)
Explanation:
\(W=\int_{1}^{2} F d x\)
Work, Energy and Power
268759
A body is displaced from \((0,0)\) to \((1 \mathrm{~m}, 1 \mathrm{~m})\) along the path \(\mathrm{x}=\mathrm{y}\) by a force \(F=x^{2} \hat{j}+y i \hat{\theta} \hat{n}\). The work done by this force will be
1 \(\frac{4}{3} J\)
2 \(\frac{5}{6} J\)
3 \(\frac{3}{2} J\)
4 \(\frac{7}{5} J\)
Explanation:
\(W=\int_{(0,0)}^{(1,1)} \vec{F} \cdot \overrightarrow{d s} ;\) Here \(\overrightarrow{d s}=d x \hat{i}+d y \hat{j}+d z \hat{k}\)
\(\therefore W=\int_{(0,0)}^{(1,1)}\left(x^{2} d y+y d x\right)=\int_{(0,0)}^{(1,1)}\left(y^{2} d y+x \cdot d x\right)\) (as \(\mathrm{x}=\mathrm{y})\)
Work, Energy and Power
268760
A particle moves under the effect of a force\(F=C x\) from \(x=0\) to \(x=x_{1}\). The work done in the process is (treat \(C\) as a constant)
1 \(\mathrm{Cx}_{1}^{2}\)
2 \(C^{2} / x_{1}^{2}\)
3 \(1 / 2 \mathrm{Cx}_{1}^{2}\)
4 \(1 / 2 C^{2} / x_{1}^{2}\)
Explanation:
\(W=\int_{0}^{X_{1}} F d x\)
Work, Energy and Power
268761
Under the action of a force, a \(2 \mathbf{k g}\) body moves such that its position ' \(x\) ' in meters as a function of time ' \(t\) ' in seconds given by:
\(x=t^{2} / 2\). The work done by the force in the first 5 seconds is
1 \(2.5 \mathrm{~J}\)
2 \(0.25 \mathrm{~J}\)
3 \(25 \mathrm{~J}\)
4 \(250 \mathrm{~J}\)
Explanation:
\(W=\frac{1}{2} m v^{2}\) where \(v=\frac{d x}{d t}\)
268644
An object has a displacement from position vector \(\vec{r}_{1}=(2 \hat{i}+3 \hat{j}) \mathbf{m}\) to \(\vec{r}_{2}=(4 \hat{i}+6 \hat{j}) \mathbf{m}\) under a force \(\vec{F}=\left(3 x^{2} \hat{i}+2 y \hat{j}\right) \mathbf{N}\), then work done by the force is
1 \(24 \mathrm{~J}\)
2 \(33 \mathrm{~J}\)
3 \(83 \mathrm{~J}\)
4 \(45 \mathrm{~J}\)
Explanation:
\(W=\int d w=\int_{x_{1}}^{x_{2}} F_{x} d x+\int_{y_{1}}^{y_{2}} F_{y} d y\)
Work, Energy and Power
268707
A force \(\mathbf{F}=(2+x) \mathbf{N}\) acts on a particle in \(x\) direction where ' \(x\) ' is in metre. The work done by this force during a displacement from \(x=1 \mathbf{m}\) to \(x=2 \mathbf{m}\) is
1 \(2 \mathrm{~J}\)
2 \(3.5 \mathrm{~J}\)
3 \(4.5 \mathrm{~J}\)
4 \(5 \mathrm{~J}\)
Explanation:
\(W=\int_{1}^{2} F d x\)
Work, Energy and Power
268759
A body is displaced from \((0,0)\) to \((1 \mathrm{~m}, 1 \mathrm{~m})\) along the path \(\mathrm{x}=\mathrm{y}\) by a force \(F=x^{2} \hat{j}+y i \hat{\theta} \hat{n}\). The work done by this force will be
1 \(\frac{4}{3} J\)
2 \(\frac{5}{6} J\)
3 \(\frac{3}{2} J\)
4 \(\frac{7}{5} J\)
Explanation:
\(W=\int_{(0,0)}^{(1,1)} \vec{F} \cdot \overrightarrow{d s} ;\) Here \(\overrightarrow{d s}=d x \hat{i}+d y \hat{j}+d z \hat{k}\)
\(\therefore W=\int_{(0,0)}^{(1,1)}\left(x^{2} d y+y d x\right)=\int_{(0,0)}^{(1,1)}\left(y^{2} d y+x \cdot d x\right)\) (as \(\mathrm{x}=\mathrm{y})\)
Work, Energy and Power
268760
A particle moves under the effect of a force\(F=C x\) from \(x=0\) to \(x=x_{1}\). The work done in the process is (treat \(C\) as a constant)
1 \(\mathrm{Cx}_{1}^{2}\)
2 \(C^{2} / x_{1}^{2}\)
3 \(1 / 2 \mathrm{Cx}_{1}^{2}\)
4 \(1 / 2 C^{2} / x_{1}^{2}\)
Explanation:
\(W=\int_{0}^{X_{1}} F d x\)
Work, Energy and Power
268761
Under the action of a force, a \(2 \mathbf{k g}\) body moves such that its position ' \(x\) ' in meters as a function of time ' \(t\) ' in seconds given by:
\(x=t^{2} / 2\). The work done by the force in the first 5 seconds is
1 \(2.5 \mathrm{~J}\)
2 \(0.25 \mathrm{~J}\)
3 \(25 \mathrm{~J}\)
4 \(250 \mathrm{~J}\)
Explanation:
\(W=\frac{1}{2} m v^{2}\) where \(v=\frac{d x}{d t}\)
268644
An object has a displacement from position vector \(\vec{r}_{1}=(2 \hat{i}+3 \hat{j}) \mathbf{m}\) to \(\vec{r}_{2}=(4 \hat{i}+6 \hat{j}) \mathbf{m}\) under a force \(\vec{F}=\left(3 x^{2} \hat{i}+2 y \hat{j}\right) \mathbf{N}\), then work done by the force is
1 \(24 \mathrm{~J}\)
2 \(33 \mathrm{~J}\)
3 \(83 \mathrm{~J}\)
4 \(45 \mathrm{~J}\)
Explanation:
\(W=\int d w=\int_{x_{1}}^{x_{2}} F_{x} d x+\int_{y_{1}}^{y_{2}} F_{y} d y\)
Work, Energy and Power
268707
A force \(\mathbf{F}=(2+x) \mathbf{N}\) acts on a particle in \(x\) direction where ' \(x\) ' is in metre. The work done by this force during a displacement from \(x=1 \mathbf{m}\) to \(x=2 \mathbf{m}\) is
1 \(2 \mathrm{~J}\)
2 \(3.5 \mathrm{~J}\)
3 \(4.5 \mathrm{~J}\)
4 \(5 \mathrm{~J}\)
Explanation:
\(W=\int_{1}^{2} F d x\)
Work, Energy and Power
268759
A body is displaced from \((0,0)\) to \((1 \mathrm{~m}, 1 \mathrm{~m})\) along the path \(\mathrm{x}=\mathrm{y}\) by a force \(F=x^{2} \hat{j}+y i \hat{\theta} \hat{n}\). The work done by this force will be
1 \(\frac{4}{3} J\)
2 \(\frac{5}{6} J\)
3 \(\frac{3}{2} J\)
4 \(\frac{7}{5} J\)
Explanation:
\(W=\int_{(0,0)}^{(1,1)} \vec{F} \cdot \overrightarrow{d s} ;\) Here \(\overrightarrow{d s}=d x \hat{i}+d y \hat{j}+d z \hat{k}\)
\(\therefore W=\int_{(0,0)}^{(1,1)}\left(x^{2} d y+y d x\right)=\int_{(0,0)}^{(1,1)}\left(y^{2} d y+x \cdot d x\right)\) (as \(\mathrm{x}=\mathrm{y})\)
Work, Energy and Power
268760
A particle moves under the effect of a force\(F=C x\) from \(x=0\) to \(x=x_{1}\). The work done in the process is (treat \(C\) as a constant)
1 \(\mathrm{Cx}_{1}^{2}\)
2 \(C^{2} / x_{1}^{2}\)
3 \(1 / 2 \mathrm{Cx}_{1}^{2}\)
4 \(1 / 2 C^{2} / x_{1}^{2}\)
Explanation:
\(W=\int_{0}^{X_{1}} F d x\)
Work, Energy and Power
268761
Under the action of a force, a \(2 \mathbf{k g}\) body moves such that its position ' \(x\) ' in meters as a function of time ' \(t\) ' in seconds given by:
\(x=t^{2} / 2\). The work done by the force in the first 5 seconds is
1 \(2.5 \mathrm{~J}\)
2 \(0.25 \mathrm{~J}\)
3 \(25 \mathrm{~J}\)
4 \(250 \mathrm{~J}\)
Explanation:
\(W=\frac{1}{2} m v^{2}\) where \(v=\frac{d x}{d t}\)
