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A block is drag on a smooth plane with the help of a rope which moves with velocity V as shown in the figure. The horizontal velocity of the block is (a) V (b) V/sin ø (c) Vsinø (d) V/cosø here theta is ø?
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Understanding the Problem
When a block is dragged on a smooth plane using a rope, the motion of the rope creates an angle θ with the horizontal. This angle affects how the block's horizontal velocity is determined.
Analyzing the Velocity Components
The velocity of the rope, V, can be broken down into its horizontal and vertical components based on the angle θ.
Key Points:
- The horizontal component of velocity (Vx) is given by Vx = V * cos(θ).
- The vertical component of velocity (Vy) is given by Vy = V * sin(θ).
Relationship Between Rope and Block
Since the block is being dragged along the plane, its horizontal velocity is influenced directly by the motion of the rope. The block will move horizontally due to the horizontal component of the rope’s velocity.
Calculating the Horizontal Velocity
To find the horizontal velocity of the block, we need to analyze how the angle θ affects the overall motion:
- The rope's velocity (V) acts at an angle, and the block's horizontal motion depends on this angle.
- Therefore, the relationship can be stated as:
Horizontal Velocity of the Block:
- V_horizontal = V * cos(θ)
However, since the block is not moving vertically, we consider the projection of the rope's velocity along the horizontal direction.
Conclusion
After analyzing the components, the correct expression for the horizontal velocity of the block is:
Answer: V / cos(θ)
Thus, the correct option is (d) V/cos(θ).
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A block is drag on a smooth plane with the help of a rope which moves with velocity V as shown in the figure. The horizontal velocity of the block is (a) V (b) V/sin ø (c) Vsinø (d) V/cosø here theta is ø?
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