2 short questions...?

Question

1. A boy rides a roller coaster. He passes over the top of a 31 m tall hill with a speed of 5.0 m/s. Assuming no friction, what will be his speed 650 feet further along the ride when he is at a height of 8.0 m above the ground?

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2. A 250 kg satellite is in a circular orbit around the earth at an altitude of 2400 km.

a. Calculate the gravitational force on the satellite.

b. Calculate the speed of the satellite.

Thank you!!

Answer 1

Yes, use the equation the first poster said. However, please, please, please use gravity as 9.8 m/s^2. Science is a metric world. Using imperial gets the Mars Lander crashed.

Answer 2

Your answers to both can be worked out using the equation s =ut +(at)^2 and gravity at 32ft/ses/sec.

Answer 3

1)  KE, Kinetic energy =  1/2*m*V*V

     PE, potential energy is  m*g*h.

      energy is conserved.

      decrease in PE = increase in KE

       m*g*( 31 - 8 ) =  1/2*m* ( V2**2 - V1**2)

        divide by m both sides.

        9.81*( 23 )  =  1/2 ( V2**2 - 5**2 )

         multiply by 2 both sides

         2* 9.81* ( 23 ) = V2**2 - 25

         58.86 + 25 =  V2**2

          83.86   =  V2**2

           V2  =  9.15751

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2)  This problem is is poorly formed.  Is the altitude from the center of the earth ?? 0r from the surface ?? Centripital acceleration is

     a =  (m *  V * V ) / R

But R is from the center of rotation, and the diameter of the earth is not given.......

      ( G*  M1 * M2 ) / R **2  = force on satellite where  

        R = 2400 +  Diameter / 2

         F = m * A        so

       ( G * M1 * M2 )                        M1 * M1 *  V**2

       -----------------------       =             -----------------

       ( 2400 + D / 2 )**2                  ( 2400 + D / 2 )

Solve for  V