please do Q4 and Q5 A mass m = 71 kg slides on a frictionless track that has a d

Question

please do Q4 and Q5

A mass m = 71 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 17.1 m and finally a flat straight section at the same height as the center of the loop (17.1 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

1)

What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track?

answe: 12.7

2)

What height above the ground must the mass begin to make it around the loop-the-loop?

answer: 42.4

4)

If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (17.1 m off the ground)?

answer: ????

5)

Now a spring with spring constant k = 18100 N/m is used on the final flat surface to stop the mass. How far does the spring compress?

answer :???

Explanation / Answer

1.

Ki+Ui=Kf+Uf, The equation mg=mv^2/4

g=v^2/r. so I take g and then

get sqrt(gr) and you will get 12.7

2.

Ki+Ui=Kf+Uf

The equation mgh=1/2mv^2=answer+(2r)
so h=1/2v^2/g and then add 2r.

=42.4

4.

Energy: ½mVt² + m*g*(2*R) = ½m*Vb² Vb² = Vt² + 4*g*R

v=28.76 m/s

Vf² = Vt² + 2*g*R Vf = 21.7 m/s

5

.Es = KE
½kx² = ½mVf² x² = m*Vf²/k
x = 1.28 m

the spring will compress 1.28 m

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