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qwelly [4]
3 years ago
10

What force does a trampoline have to apply to a gymnast to accelerate her straight up at ? Note that the answer is independent o

f the velocity of the gymnastâshe can be moving either up or down, or be stationary.
Physics
1 answer:
Andrew [12]3 years ago
5 0

Answer: Force applied by trampoline = 778.5 N

<em>Note: The question is incomplete.</em>

<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary. </em>

Explanation:

The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.

= (m * a) + (m * g)

= m ( a + g)

= 45 kg ( 7.50 *  9.80) m/s²

Force applied by trampoline = 778.5 N

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
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Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

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