- 1(current)
Q. No.A man of mass \(M\) throws a ball of mass \(m\) vertically upward. At the beginning of the throw, he holds the ball at rest, and releases it at height \(h\) above at the end of it. The ball travels up to a maximum height \(H\) above the point of release. The work done by the man on the ball is \(W_1\) and that done by gravity on the ball is of magnitude \(W_2.\) Both \(W_1\) and \(W_2\) are the work done during the entire motion — from when the man begins the throw, till the ball reaches its maximum height. Then:
| 1. | \(\dfrac{W_1}{W_2}=\dfrac mM\) | 2. | \(\dfrac{W_1}{W_2}=\dfrac hH\) |
| 3. | \(\dfrac{W_1}{W_2}=\dfrac{h}{h+H}\) | 4. | \(\dfrac{W_1}{W_2}=\dfrac 11\) |
Subtopic: Work done by constant force |
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A block of mass \(m\) is slid slowly up a smooth incline \(AB\) and then, its is slowly slid down a second smooth incline \(BC\) until it reaches the ground. Let the magnitude of work done by the person sliding the block be \(W_1\) for the upward path \((AB)\) and \(W_2\) for the downward path \((BC).\) Then,
1. \(W_1=W_2\)
2. \(W_1\text{cos}30^\circ=W_2\text{cos}60^\circ\)
3. \(W\text{sin}30^\circ=W_2\text{sin}60^\circ\)
4. \(W_1\text{tan}30^\circ=W_2\text{tan}60^\circ\)
1. \(W_1=W_2\)
2. \(W_1\text{cos}30^\circ=W_2\text{cos}60^\circ\)
3. \(W\text{sin}30^\circ=W_2\text{sin}60^\circ\)
4. \(W_1\text{tan}30^\circ=W_2\text{tan}60^\circ\)
Subtopic: Work done by constant force |
From NCERT
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A constant additional force acts on a body thrown vertically upward under the earth's gravity, the force always opposing the motion. The magnitude of work done by the force during the upward motion is \(W_1\) and during the downward motion is \(W_2.\) Then,
| 1. | \(W_1>W_2\) |
| 2. | \(W_1<W_2\) |
| 3. | \(W_1=W_2\) |
| 4. | Any of the above may be true depending on the initial speed of the body |
Subtopic: Work done by constant force |
58%
From NCERT
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