Question

How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track

Answer 1

Answer:

Check attachment for complete questions, the question is not complete

Explanation:

Check attachment for solution

Answer 2

Complete Question

The complete Question is shown on the first and second uploaded image

Answer:

The speed at which they need to push the mass is v = 13.1 m/s

Explanation:

In order to solve this problem we need to consider conservation of energy when the block is at the top of the inclined plane and also when it is on top of the loop

Now Applying the law of conservation of energy

        $mg (2R) + \frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2 + mg(2R)$

  where   $mg (2R)$ is potential energy and $\frac{1}{2} mv^2$ is kinetic energy

  and $v_{top}$ is the velocity at the top inclined plane and the top of the loop

        Now considering the formula

                           $\frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2$

                            $v^2 = v_{top}^2$

                            $v = v_{top}$

Now to obtain $v_{top}$

   Looking at the question we can say that the centripetal force that made the block move around loop without leaving the track is q=equivalent to the centripetal force  so we have

            $mg = \frac{mv_{top}^2}{R}$

The m would cancel out each other then cross- multiplying

             $gR = v^2_{top}$

         $v_{top} = \sqrt{gR}$

                 $= \sqrt{(9.8 m/s^2)(17.4\ m)}$

                $= 13.05 m/s$

                $\approx 13.1 m/s$