A box of m = 10.0 (kg) rests on a frictionless ramp set at...

A box of m = 10.0 (kg) rests on a frictionless ramp set at an angle of ?1 = 30 degrees, and rests at a height of h1 = 2.88 (m). The distance between the first ramp and the second ramp d1= 5.0 (m) has a kinetic coefficient of friction ?k= 0.3. The second frictionless ramp is situated at an angle ?2= 20 degrees. a) What is the velocity of the box just as it reaches the floor? (Treat the box as a point mass that starts at h1 so that you do not need to consider parts of the box being on the ramp and parts of it being on the floor.) b) What is the velocity of the box when it reaches the second ramp? c) How far up the second ramp d2 does the box slide? (Bear in mind what I am asking for here, I want the distance up the ramp not just the height.)

Transcript

00:01 In the question, a box of mass 10 kilogram is a rest on a frictionless ram at an angle of 30 degrees.

00:07 And the height is 2 .88 meters.

00:10 The distance between the first and second ramp is 5 meters and the coefficient of kinetic friction is 0 .3.

00:15 And the second one is at 20 degrees.

00:17 So we need to find a velocity just as it reaches the floor.

00:21 And the velocity just as it reaches the second ram and the velocity and the, sorry, how far as in the distance, it travels along the second slide.

00:32 So this is completely based on work energy theorem.

00:35 So let's say this is v1 and this is v2.

00:38 So for finding part a, we are going to conserve the energy, that initial energy should be equal to the final energy.

00:46 Initial is only the potential, so that's mgh1.

00:50 And the final is only the kinetic, so that's half mv1 square because that came on reference of this we take as reference, so there's no potential.

00:59 M &m is cancer.

01:00 If you multiply both sides by two, we have 2gh1 is equal to v1 square.

01:06 If you flip the equation, then we have v1 square as 2gh1.

01:11 So the value of v1 by taking square root both sides comes out as root of 2gh1.

01:18 G is of course 9 .8h1 is 2 .88.

01:26 So this value comes out as 2 times 9 .8 times 2 .8 times 2 .8.

01:31 8 8 square root of this value comes out as 7 .51 meter per second.

01:39 So that is the answer to part a.

01:42 In part b, it's asking the velocity as it reaches the second ramp.

01:46 So again, the work energy theorem, the energy at the point, let's call it a, plus the work done by friction is equal to the energy, which is at b.

02:02 Remember, the potential energy is not changing, only the kinetic is changing.

02:06 So half m v1 square plus work done by the friction is f .d1 and f dot d1 and is equal to half mv2 square.

02:18 So i think v1 squares already 2gh1.

02:23 In fact, we just square it up to get the value as 56 .448.

02:28 So half m 56 .448.

02:33 Now f dot d means there's a dot product and friction is over here and the displacement is over on the other side...

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