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KatRina [158]
3 years ago
6

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 1/2 mv2

Physics
2 answers:
snow_lady [41]3 years ago
7 0

Explanation :

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by \dfrac{1}{2}mv^2.

We know that work done is equal to the change in kinetic energy.

                     W=\Delta KE

Let kinetic energy increases from 0 to \dfrac{1}{2}mv^2

 So,                F.s=\Delta KE

                       F.s=\dfrac{1}{2}mv^2........................(1)

According to second condition if both force and distance is tripled, then kinetic energy increases by KE^'.

So,                        3F\times 3s=\Delta KE^'

                               9F.s=\Delta KE'..........................(2)

Using equation (1) in (2)

\dfrac{9}{2}mv^2=\Delta KE^'

This is the resulting increase in kinetic energy.

Original increase in kinetic energy is \dfrac{9}{2}mv^2.

agasfer [191]3 years ago
6 0
W work
F force 
s distance

If F = constant:

W₁ = F·s

If you triple the force and the distance:

W₂ = 3F · 3s = 9 F·s = 9 W₁

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Answer:

The height is  h_c = 42.857

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   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

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

  The angular velocity w is mathematically represented as

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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

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             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

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                 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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∅ = 89.44°

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Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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