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A Small Block of Mass 200 G is Kept at the Top of a Frictionless Incline Which is 10 M Long and 3⋅2 M High. How Much Work Was Required (A) to Lift the Block from the Ground and Put It an the Top, - Physics

Sum

A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically to the ground (d) it slides down the incline? Take g = 10 m/s2

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Solution

\[\text{ Given }, \]

\[\text{ Mass of the block, m = 200 g = 0 . 2 kg} \]

\[\text{ Length of the incline, s = 10 m }, \]

\[\text{ Height of the incline, h = 3 . 2 m } \]

\[\text{ Acceleration due to gravity, g = 10 m/ s}^2\]

(a) Work done, W = mgh = 0.2 × 10 × 3.2 =6 .4 J

(b)  Work done to slide the block up the incline

\[\text{ W }= \left( \text{ mg }  \sin \theta \right) \times \text{ s } \]

\[ = \left( 0 . 2 \right) \times 10 \times \left( 3 . 2/10 \right) \times 10\]

\[ = 6 . 4 \text{ J } \]

(c) Let final velocity be v when the block falls to the ground vertically.
Change in the kinetic energy = Work done

\[\frac{1}{2}\text{mv}^2 - 0 = 6 . 4 \text{ J }\]

\[\Rightarrow \nu = 8 \text{ m/s }\]

(d) Let \[\nu\] be the final velocity of the block when it reaches the ground by sliding.

\[\frac{1}{2}m \nu^2 - 0 = 6 . 4 \text{ J }\]
\[\Rightarrow \text{ v = 8  m/s } \]
  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 8 Work and Energy
Q 34 | Page 134
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