355783
A body, constrained to move in the \(y\) - direction , is subjected to a force \(F=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) N\).What is the work done by this force in moving the body through a distance of \(10 \mathrm{~m}\) along the \(y\) axis?
1 20 \(J\)
2 150 \(J\)
3 160 \(J\)
4 190 \(J\)
Explanation:
\(W=\) (\(y\) component of force) \(x\) ( displacement along \(y\) - axis) \(=15 \times 10=150 \mathrm{~J}\)
PHXI06:WORK ENERGY AND POWER
355784
A force \(\vec F = (5\hat i + 3\hat j + 2\hat k)N\) is applied over a particle which displaces it from its origin to the point\(\vec r = (2\hat i - \hat j)m\). The work done on the particle in joule is
355785
A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and third second of the motion of ball is
1 \(1: 2: 3\)
2 \(1: 4: 16\)
3 \(1: 3: 5\)
4 \(1: 9: 25\)
Explanation:
Ratio of displacement is \(\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(3)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}\) or \(1: 3: 5\) Therefore ratio of work done will be also be \(1: 3: 5\)
PHXI06:WORK ENERGY AND POWER
355786
A force acting on a particle is \((2 \hat{i}+3 \hat{j}) N\). Work done by the force is zero, when a particle is moved on the line \(3y + kx = 5\). Here value of \(k\) is
1 2
2 4
3 6
4 8
Explanation:
Line and force should be perpendicular, as the work done is given to be zero \({m_1} = \frac{{{F_y}}}{{{F_x}}} = \frac{3}{2} = {\rm{ }}\) slope of force \({m_2} = - \frac{k}{3} = \) slope of line \({m_1}\;{m_2} = - 1\) Therefore,\(k = 2\)
355783
A body, constrained to move in the \(y\) - direction , is subjected to a force \(F=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) N\).What is the work done by this force in moving the body through a distance of \(10 \mathrm{~m}\) along the \(y\) axis?
1 20 \(J\)
2 150 \(J\)
3 160 \(J\)
4 190 \(J\)
Explanation:
\(W=\) (\(y\) component of force) \(x\) ( displacement along \(y\) - axis) \(=15 \times 10=150 \mathrm{~J}\)
PHXI06:WORK ENERGY AND POWER
355784
A force \(\vec F = (5\hat i + 3\hat j + 2\hat k)N\) is applied over a particle which displaces it from its origin to the point\(\vec r = (2\hat i - \hat j)m\). The work done on the particle in joule is
355785
A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and third second of the motion of ball is
1 \(1: 2: 3\)
2 \(1: 4: 16\)
3 \(1: 3: 5\)
4 \(1: 9: 25\)
Explanation:
Ratio of displacement is \(\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(3)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}\) or \(1: 3: 5\) Therefore ratio of work done will be also be \(1: 3: 5\)
PHXI06:WORK ENERGY AND POWER
355786
A force acting on a particle is \((2 \hat{i}+3 \hat{j}) N\). Work done by the force is zero, when a particle is moved on the line \(3y + kx = 5\). Here value of \(k\) is
1 2
2 4
3 6
4 8
Explanation:
Line and force should be perpendicular, as the work done is given to be zero \({m_1} = \frac{{{F_y}}}{{{F_x}}} = \frac{3}{2} = {\rm{ }}\) slope of force \({m_2} = - \frac{k}{3} = \) slope of line \({m_1}\;{m_2} = - 1\) Therefore,\(k = 2\)
355783
A body, constrained to move in the \(y\) - direction , is subjected to a force \(F=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) N\).What is the work done by this force in moving the body through a distance of \(10 \mathrm{~m}\) along the \(y\) axis?
1 20 \(J\)
2 150 \(J\)
3 160 \(J\)
4 190 \(J\)
Explanation:
\(W=\) (\(y\) component of force) \(x\) ( displacement along \(y\) - axis) \(=15 \times 10=150 \mathrm{~J}\)
PHXI06:WORK ENERGY AND POWER
355784
A force \(\vec F = (5\hat i + 3\hat j + 2\hat k)N\) is applied over a particle which displaces it from its origin to the point\(\vec r = (2\hat i - \hat j)m\). The work done on the particle in joule is
355785
A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and third second of the motion of ball is
1 \(1: 2: 3\)
2 \(1: 4: 16\)
3 \(1: 3: 5\)
4 \(1: 9: 25\)
Explanation:
Ratio of displacement is \(\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(3)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}\) or \(1: 3: 5\) Therefore ratio of work done will be also be \(1: 3: 5\)
PHXI06:WORK ENERGY AND POWER
355786
A force acting on a particle is \((2 \hat{i}+3 \hat{j}) N\). Work done by the force is zero, when a particle is moved on the line \(3y + kx = 5\). Here value of \(k\) is
1 2
2 4
3 6
4 8
Explanation:
Line and force should be perpendicular, as the work done is given to be zero \({m_1} = \frac{{{F_y}}}{{{F_x}}} = \frac{3}{2} = {\rm{ }}\) slope of force \({m_2} = - \frac{k}{3} = \) slope of line \({m_1}\;{m_2} = - 1\) Therefore,\(k = 2\)
355783
A body, constrained to move in the \(y\) - direction , is subjected to a force \(F=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) N\).What is the work done by this force in moving the body through a distance of \(10 \mathrm{~m}\) along the \(y\) axis?
1 20 \(J\)
2 150 \(J\)
3 160 \(J\)
4 190 \(J\)
Explanation:
\(W=\) (\(y\) component of force) \(x\) ( displacement along \(y\) - axis) \(=15 \times 10=150 \mathrm{~J}\)
PHXI06:WORK ENERGY AND POWER
355784
A force \(\vec F = (5\hat i + 3\hat j + 2\hat k)N\) is applied over a particle which displaces it from its origin to the point\(\vec r = (2\hat i - \hat j)m\). The work done on the particle in joule is
355785
A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and third second of the motion of ball is
1 \(1: 2: 3\)
2 \(1: 4: 16\)
3 \(1: 3: 5\)
4 \(1: 9: 25\)
Explanation:
Ratio of displacement is \(\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(3)^{2}-\dfrac{1}{2} \mathrm{~g}(1)^{2} \dfrac{1}{2} \mathrm{~g}(2)^{2}\) or \(1: 3: 5\) Therefore ratio of work done will be also be \(1: 3: 5\)
PHXI06:WORK ENERGY AND POWER
355786
A force acting on a particle is \((2 \hat{i}+3 \hat{j}) N\). Work done by the force is zero, when a particle is moved on the line \(3y + kx = 5\). Here value of \(k\) is
1 2
2 4
3 6
4 8
Explanation:
Line and force should be perpendicular, as the work done is given to be zero \({m_1} = \frac{{{F_y}}}{{{F_x}}} = \frac{3}{2} = {\rm{ }}\) slope of force \({m_2} = - \frac{k}{3} = \) slope of line \({m_1}\;{m_2} = - 1\) Therefore,\(k = 2\)
