A force of 5 N making an angle $\theta$ with the horizontal acting on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1 J then the component of the force is:

 1 1.5 N 2 2.5 N 3 3.5 N 4 4.5 N
Subtopic:  Work Energy Theorem |
87%
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The bob of a simple pendulum having length l, is displaced from the mean position to an angular position θ with respect to vertical. If it is released, then the velocity of the bob at the lowest position will be:

1. $\sqrt{2gl\left(1-\mathrm{cos}\theta \right)}$

2. $\sqrt{2gl\left(1+\mathrm{cos}}$ $\theta \right)$

3. $\sqrt{2gl}$ $\mathrm{cos}\theta$

4. $\sqrt{2gl}$

Subtopic:  Work Energy Theorem |
79%
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AIPMT - 2000
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A block is carried slowly up an inclined plane. If ${\mathrm{W}}_{\mathrm{f}}$ is work done by the friction, ${\mathrm{W}}_{\mathrm{N}}$ is work done by the reaction force, ${\mathrm{W}}_{\mathrm{g}}$ is work done by the gravitational force and ${\mathrm{W}}_{\mathrm{ex}}$ is the work done by an external force, then choose the correct relation(s):

1.  ${\mathrm{W}}_{\mathrm{N}}$ $+$ ${\mathrm{W}}_{\mathrm{f}}$ $+$ ${\mathrm{W}}_{\mathrm{g}}$ $+$ ${\mathrm{W}}_{\mathrm{ex}}$ $=$ $0$

2.  ${\mathrm{W}}_{\mathrm{N}}$ = 0

3.  ${\mathrm{W}}_{\mathrm{ex}}$ $+$  ${\mathrm{W}}_{\mathrm{f}}$ $=$ $-{\mathrm{W}}_{\mathrm{g}}$

4.  All of these

Subtopic:  Work Energy Theorem |
69%
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A body of mass 'm' is released from the top of a fixed rough inclined plane as shown in the figure. If the frictional force has magnitude F, then the body will reach the bottom with a velocity: $\left(\mathrm{L}=\sqrt{2}\mathrm{h}\right)$

 1 $$\sqrt{2 g h}$$ 2 $$\sqrt{\frac{2 F h}{m}}$$ 3 $$\sqrt{2 g h+\frac{2 F h}{m}}$$ 4 $$\sqrt{2 g h-\frac{2 \sqrt{2} F h}{m}}$$
Subtopic:  Work Energy Theorem |
62%
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A block is released from rest from a height of h = 5 m. After travelling through the smooth curved surface, it moves on the rough horizontal surface through a length l = 8 m and climbs onto the other smooth curved surface at a height h'. If $\mathrm{\mu }$ = 0.5, find h'.

 1 2 m 2 3 m 3 1 m 4 Zero
Subtopic:  Work Energy Theorem |
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A block of mass 'm' is connected to a spring of force constant K. Initially, the block is at rest and the spring is relaxed. A constant force F is applied horizontally towards the right. The maximum speed of the block will be:

 1 $\frac{\mathrm{F}}{\sqrt{2\mathrm{mK}}}$ 2 $\frac{\sqrt{2}\mathrm{F}}{\sqrt{\mathrm{mK}}}$ 3 $\frac{\mathrm{F}}{\sqrt{\mathrm{mK}}}$ 4 $\frac{2\mathrm{F}}{\sqrt{2\mathrm{mK}}}$
Subtopic:  Work Energy Theorem |
57%
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