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Strings stretched horizontally on a rough surface

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.

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Strings stretched horizontally on a rough surface

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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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