1) At what height h above the ground is the block released? Answer in units of m

Question

1) At what height h above the ground is the block released? Answer in units of m

2) What is the the speed of the block when itleaves the track?Answer in units of m/s

3) What is the total speed of the block when ithits the ground?Answer in units of m/s

 

A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track (as shown in the ?gure below). It leaves the track horizontally, ?ies through the air, and subsequently strikes the ground.The acceleration of gravity is 9.81 m/s2. 1) At what height h above the ground is the block released? Answer in units of m 2) What is the the speed of the block when itleaves the track?Answer in units of m/s 3) What is the total speed of the block when ithits the ground?Answer in units of m/s

Explanation / Answer

it travels a distance (h-2.3) vertically before coming on the rough horizontal area
so its speed will be
v2 = u2 +2as
v2 = 2g(h-2.3)

when it be on rough horizontal area a friction force will act in opposite direction of its motion

F = Mg =0.2Mg

so its as a deacceleration of 0.2g.

its speed after traveling rough section of 1.4 m

vf2 = v2 + 2(-0.2g)1.4

vf2 = 2g(h-2.3) -0.56g = 2gh - 5.16g

vf = (2gh-5.16g) m/s

with this speed vf block leaves that track now it has to go 4.72m horizontaly with the speed vf and

2.3 vertically with g .

2.3 = (1/2)gt2

so t = 4.6/g sec

in this time its travels also 4.72m horizontal os

4.72 = (2gh-5.16g) 4.6/g

22.28 = (2gh-5.16g)(4.16/g)

2h - 5.16 = 5.36

h = 10.52/2 = 5.26 m ...................Ans(1)

vf = (2*9.81*5.26 - 5.16*9.81) = 7.25 m/s    ............... Ans(2)

when its on ground it has two component one is horizontal that is 7.25 m/s and another is vertival

which is gt = 9.81*4.6/g = 9.81*0.685 = 6.72 m/s

so total speed will be (7.252 + 6.722 ) = 9.885m/s ...........Ans(3)

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