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Margaret [11]
2 years ago
13

A box, initially at rest, has 36.7 N of force exerted on it for 2.81 s. If the box has a mass of 7.41 kg, what was its velocity

at this time?
Physics
2 answers:
ElenaW [278]2 years ago
8 0

Answer:

13.91 m/s

Explanation:

First we need to find the acceleration:

Acceleration = Force/mass

Acceleration = 36.7N/7.41 kg

Acceleration = 4.95 m/s² (rounded to two decimal places)

Then we find the velocity:

Velocity = Acceleration * Time

Velocity = 4.95 m/s² * 2.81 s

Velocity = 13.91 m/s (rounded to two decimal places)

Oksanka [162]2 years ago
7 0

Hi there!

We can begin by calculating the acceleration of the box using Newton's Second Law:

\large\boxed{\Sigma F = ma}

∑F = Net force (N)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

\frac{\Sigma F}{m} = a\\\\\frac{36.7}{7.41} = 4.95 m/s^2

Now, we can use the kinematic equation:

v_f = v_0 + at

vf = Final velocity (? m/s)

v0 = initial velocity (0 m/s, from rest)

a = acceleration (m/s²)

t = time (s)

Plug in the values:

v_f = 0 + 4.95(2.81)\\\\v_f = \boxed{13.91 m/s}

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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 )

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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

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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