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Calculation of Distance Traveled by a Ball on a Rough Surface with Stretched Strings

A 2kg ball is attached to a 2m string with a tension of 0.40N, fixed to a wall, and pulled to 2.8m. The rough surface has a coefficient of friction (m) of 0.3. The scenario examines the distance the ball travels after being released, applying principles of energy conservation and frictional work. Initial energy and work done against friction are calculated to derive the final velocity. The analysis involves understanding how friction affects the ball's motion after release, ultimately revealing the distance traveled before stopping.

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Calculation of Distance Traveled by a Ball on a Rough Surface with Stretched Strings

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  1. Strings stretched horizontally on a rough surface A ball of mass 2kg is attached to a 2m string with = 40N. One end of the string is attached to a wall and the ball is then pulled out so that the length of the string is 2.8m. The surface is rough with m = 0.3. Find the distance travelled by the ball after it is released. g = 10ms–2 0.8m 2m N A S

  2. Strings stretched horizontally on a rough surface A ball of mass 2kg is attached to a 2m string with = 40N. One end of the string is attached to a wall and the ball is then pulled out so that the length of the string is 2.8m. The surface is rough with m = 0.3. Find the distance travelled by the ball after it is released. g = 10ms–2 0.8m 2m N A S

  3. 2m 0.8m N A S

  4. 2m 0.8m N A S Energy at start – work done overcoming the resistance = energy at N (1) Work done overcoming friction = Fr× distance moved = mR×0.8 = 0.3×2g×0.8 = 4.8 J So using (1) 6.4 – 4.8 = v2 v2 = 1.6 v = 1.26ms–1

  5. 2m 0.8m N A S If the ball stops at S Work done overcoming friction = Fr× distance moved = mR×x = 0.3×2g×x = 0.6gx J At S KE = 0 EPE = 0 Energy at start – work done overcoming the resistance = energy at N (1)

  6. x 1–x 1m 1m Two strings stretched horizontally on a rough surface with a mass in the centre Two 1m strings with l = 20N and 30N respectively are tied and then stretched between two walls 3m apart. A ball of mass 2kg is tied to the centre and then pulled 0.2m to the right. The surface is rough : m = 0.1. Find where the ball first comes to rest E A In the equilibrium position the extension of each side is x and 1 – x respectively When the ball is pulled 0.2m to the right the extension on the left is x + 0.2m and on the right is 1 – x – 0.2m i.e 0.8 – x

  7. x 1–x 1m 1m E A = 0.6m

  8. 1m 1m E A 0.6+0.2 0.4 –0.2

  9. 1m 1m 1m 1m E E A A 0.8 0.2 0.2+y 0.8–y Suppose the ball now travels a distance y: the extensions will now be 0.8 – y and 0.2 + y respectively Energy at start – work done overcoming the resistance = energy at S (1) • Work done overcoming friction = Fr× distance moved = mR×y = 0.1×2g×y = 2y

  10. 1m 1m 1m 1m E E A A 0.8 0.2 0.2+y 0.8–y Energy at start – work done overcoming the resistance = energy at S (1) So it first stops after travelling 0.32m from A i.e 1 + 0.8 – 0.32 from the left wall = 1.48m

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