A mass m = 73 kg slides on a frictionless track that has a drop, followed by a l

Question

A mass m = 73 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 18.1 m and finally a flat straight section at the same height as the center of the loop (18.1 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?

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

3)If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

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 (18.1 m off the ground)?

5)Now a spring with spring constant k = 15300 N/m is used on the final flat surface to stop the mass. How far does the spring compress?

6)The work done by the normal force on the mass (during the initial fall) is:

a)positive

b)zero

c)negative

A mass m = 73 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 18.1 m and finally a flat straight section at the same height as the center of the loop (18.1 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? 2)What height above the ground must the mass begin to make it around the loop-the-loop? 3)If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? 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 (18.1 m off the ground)? 5)Now a spring with spring constant k = 15300 N/m is used on the final flat surface to stop the mass. How far does the spring compress? 6)The work done by the normal force on the mass (during the initial fall) is: a)positive b)zero c)negative

Explanation / Answer

the centrifugal force must just cancel the gravity force. m*v^2/r = m*g
v^2/r = g = 9.8m/s^2
v^2 = 17.9*9.8 = 175.42
v = 13.24 m/s
how far would a block have to fall to gain this speed?
position = X = X0 + V0*T + 1/2A*T^2
X0 = 0, V0 = 0
X = 0.5*a*T^2
velocity = dX/dT = at =9.8*T
13.24/9.8 = T = 1.35 sec
avg speed = Vmax/2 therefore distance = 6.62m/s * 1.35 sec = 8.95M
the block must start 8.95M above the top of the loop or 35.8 +8.95 = 44.75M above the ground

the block will have 44.75M * 73kg*9.8M/s^2 = 32.014KJ of kinetic energy
K.E. = 0.5*m*V^2 = 0.5 * 73 * v^2 = 32.014KJ
V^2 = 877.1
V = 29.62M/s

at the final flat, we have to subtract the potential energy of a 17.9M rise from its' kinetic energy
32014 - 17.9*73*9.8 = 19208J
0.5*73*V^2 = 19208 V^2 = 22.94M/s

the potential energy in a spring is 0.5 * K * X^2 = 19208J
X^2 = 19208/0.5K = 2.48 in^2
X = 1.57 in

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