I have had a problem understanding these quesitons for my homework. I have previ

Q: I have had a problem understanding these quesitons for my homework. I have previ

let v2 is the speed at the top centripetal acceleration, a = v2^2/R g = v2^2/R v2 = sqrt(g*R) intial mechanical energy(lowest point) = final mechnical energy(top point) 0.5*m*vi^2 = 0.5*m*v2^2 + m*g*h 0.5*m*vi^2 = 0.5*m*v2^2 + m*g*2*R 0.5*vi^2 = 0.5*(g*R) + 2*g*R vi^2 = g*R + 4*g*R vi = sqrt(5*g*R) = sqrt(5*9.8*5.3) = 16.115 m/s

ID: 3894030 • Letter: I

Question

I have had a problem understanding these quesitons for my homework. I have previously posted this question about 3 times and they always get the incorrect answer, so i don't know why. I would really appreciate any help with these problems. Thank you!

Consider the rock and string in the figure below. The string is fastened to a hinge that allows it to swing completely around in a vertical circle. If the rock starts at the lowest point on this circle and is given an initial speed

what is the smallest value of

that will allow the rock to travel completely around the circle without the string becoming slack at the top? Assume the length of the string is 5.3 m.

Previous answers provided that were INCORRECT: 12.5 m/s and 7.21 m/s

A block is dropped onto a spring with k = 29 N/m. The block has a speed of 3.5 m/s just before it strikes the spring. If the spring compresses an amount 0.10 m before bringing the block to rest, what is the mass of the block?

Previous answers provided that were INCORRECT: .0236 kg

No previous answers were given for this problem because nobody attempted to solve and i'm really confused on this one.

I would really appreaciate any help. I will rate the best answer. Thank you!

Explanation / Answer

let v2 is the speed at the top

centripetal acceleration, a = v2^2/R

g = v2^2/R

v2 = sqrt(g*R)

intial mechanical energy(lowest point) = final mechnical energy(top point)

0.5*m*vi^2 = 0.5*m*v2^2 + m*g*h

0.5*m*vi^2 = 0.5*m*v2^2 + m*g*2*R

0.5*vi^2 = 0.5*(g*R) + 2*g*R

vi^2 = g*R + 4*g*R

vi = sqrt(5*g*R)

= sqrt(5*9.8*5.3)

= 16.115 m/s <<<<<<<<<<<<<<---------Answer

2) initial mechanical enrgy = final mechanical enrgy

m*g*x + 0.5*m*v^2 = 0.5*k*x^2

m*9.8*0.1 + 0.5*m*3.5^2 = 0.5*29*0.1^2

m*0.98 + m*6.125 = 0.145

m = 0.145/(0.98+6.125)

= 0.0204 kg

= 20.4 grams

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I have had quite a bit of trouble with this assignment for the past 11 days now.

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I have had a problem understanding these quesitons for my homework. I have previ

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