The 2.0 kg block slides down a frictionless curved ramp,...

The 2.0 kg block slides down a frictionless curved ramp, starting from rest at a height of $h = 2.0$ m. The block then slides $d = 10.0$ m on a rough horizontal surface before coming to rest. (a) What is the speed of the block at the bottom of the ramp? m/s (b) What is the energy dissipated by friction? J (c) What is the coefficient of friction between the block and the horizontal surface?

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00:01 For the given question.

00:01 In this question a block of mass that is m is equals to 2 kg is inclined on a horizontal surface having the height is equals to 2 meter.

00:13 The block inclined on the rough surface at a distance of 10 meter before it come to the test.

00:19 So first of all we have to calculate the speed of the block.

00:23 So when we use a conservation mechanical energy to determine the speed of the block the gravitational potential energy is equals to kinetic energy that is mgh equals to half mv square.

00:54 M2m cancel out so the value of velocity that is v square equals to 2gh or we can say v is equals to under root of 2gh.

01:05 On substituting the value of g and h we get 2 into g that is 9 .8 and the value of height is 2.

01:13 On solving this we get the value of velocity that is v is equals to 6 .260 meter per second.

01:22 This is the required value of the speed of the block at the bottom of the plan.

01:30 Now moving to the next part.

01:34 In part b we have to calculate the energy dissipated.

01:40 So the energy dissipated is equals to that is kinetic energy is equals to half mv final square minus half mv initial square.

02:09 On solving this we get half m common v final square minus v of initial square.

02:16 The value of final that is v final at a body has come to rest so it is zero...

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