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A 46 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above...

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lala2pretty | Student, Undergraduate | (Level 1) Honors

Posted November 18, 2009 at 10:17 PM via web

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A 46 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1 m/s. The acceleration of gravity is 9.8m/s^2.

Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Answer in units of m.

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krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted November 18, 2009 at 11:04 PM (Answer #1)

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To solve this problem we have to assume that the kinetic energy of the vaulter while running is converted partially in to potential energy for raising the vaulter to above the bar. Rest of energy is retained as kinetic energy in form of speed when the vaulter is above the bar.

Given:

Mass of vaulter = 46 kg

Running velocity/speed = v1 = 11 m/s

Velocity/speed over bar = v2 = 1 m/s

Acceleration due to gravity = g = 9.8 m/s^2

We know: Kinetic energy = (mass*velocity^2)/2

Therefore kinetic energy while running = k1

= (m*v1^2)/2 = (46*11^2)/2 = 23*121 J

And kinetic energy over the bar = k1

= (m*v2^2)/2 = (46*1^2)/2 = 23 J

Energy used for raising the vaulter to the altitude (h) above  the bar = k3

= k1 - k2 = (23*121 - 23) = 2760 J

k3 is converted in potential energy of vaulter when at height h. This can be calculated as:

Mass*(Acceleration due to gravity)*height = m*g*h = 46*9.8*h = 450.8*h J

As this is equal to k3 = k1 - k2:

450.8*h = 2760

therefore h = 2760/450.8= 6.1224 m

Answer: Altitude of vaulter when crossing the bar is 6.1224 m

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 19, 2009 at 12:57 AM (Answer #2)

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We use the law conservation of energy to find the solution.

The law of conservation of energy says that the sum of the potential energy at any point of the motion is same. Therefore, if m is the mass of the body (pole vaulter here) , his potential energy at ground level is zero and at a height, h above the ground  the PE = mgh.

His ground velocity is x and the velocity at the height, h  is y, the  kinetic energy at gound is (1/2)mx^2 and at the  height  y is (1/2)my^2. By conservation of energy:

PE+KE of the vaulter at ground =  0 + (1/2)mx^2  = constant k

PE+KE at of the vaulter at height h, = mgh + (1/2)my^2 = constant k

Therefore, from the above two equations and by the lae of conservation of energy:

(1/2)mx^2 = mgh+(1/2)my^2 or

x^2 = 2gh + y^2 , given the ground speed , x=11m/s , speed ar the height h, is y =  1m, we get:

11^2= 2*9.81*h + 1^2. So,

h = (11^2 - 1)/(2*9.81) =  6.1162 meter is the bar height.

 

 

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