1) a = g sin?1. Assuming that Adam starts from rest at the top of the ramp and t

Question

1) a = g sin?1. Assuming that Adam starts from rest at the top of the ramp and that you ignore the effect of friction, how fast would he be moving when he reaches the bottom of the ramp

2) In the space below, draw a careful free-body diagram for Adam when he is partway down the ramp. Include sliding friction –label all the forces.

3)Use your free body diagram (and Newton’s 2 law) to derive Adam’s acceleration down the ramp (Assume coefficient of kinetic friction ?k)

4) In the Mythbusters show, Adam suggests that Jamie may go faster down the slide because Jamie weighs more than Adam. Based on your result in #2, do you agree?

5) Suppose Adam starts from rest, slides a distance L1 to the bottom of the ramp, endingat a known (measured!) speed v. Which expression tells us the coefficient of kinetic friction? Present your reasoning–show the algebra. Work carefully, it’s a couple of steps.

6) In the Mythbusters show, Adam suggests that Jamie may go faster down the slide because Jamie weighs more than Adam. Based on your result in #2, do you agree?

Explanation / Answer

1) using vf^2 - vi^2 = 2(a)(d)

vf^2 - 0 = 2(9.8 sin24)(50)

vf^2 = 398.60

vf = 19.96 m/s

3) perpendicular to the ramp:

N - mgcos@1 = 0

N = mgcos@1

and frictional force, f1 =uk N = uk mg cos@1

along the ramp,

Fnet = ma

mgsin@1 - ukmgcos@1 = ma

a = gsin@1 - ukgcos@1

4) As we can see aceleration is independent of mass.

hence both will go down with same speed .

5) now using vf^2 - vi^2 = 2ad

v^2 - 0 = 2aL1

a = v^2 / 2L1

and a = gsin@1 - ukgcos@1 = v^2 / 2L1

ukgcos@1 = gsin@1 - (v^2 / 2L1)

uk = 2tan@1 - (v^2 / 2gL1cos@1 )

6. No, independent of mass.

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