Question

A block A having a mass ma = 3kg is released from rest at the position P shown and slides freely down
Center of Mass
the smooth fixed inclined ramp. When it reaches the bottom of the ramp it slides horizontally onto
the
surface of a cart of mass me = 2kg for which the coefficient of friction between the cart and the blockie
If h = 6m be the initial height of A, determine the position 'X' (with respect to cart) of the box on
the cart after it comes to rest relative to cart. (The cart moves on smooth horizontal surface.)
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Answer 1

Answer:

5

Explanation:

this is the correct answer my dear students

Answer 2

Question

A block A having a mass $m_{a} = 3kg$ is released from rest at the position P shown and slides freely down the smooth fixed inclined ramp. When it reaches the bottom of the ramp it slides horizontally onto the surface of a cart of mass $m_{c} = 2kg$ for which the coefficient of friction between the cart and the block is $\mu=\frac{2}{5}$. If h = 6m be the initial height of A, determine the position 'X' (with respect to cart) of the box on the cart after it comes to rest relative to cart. (The cart moves on smooth horizontal surface.)

Answer:

The position of the block on the cart after it comes to rest relative to cart is 117m.  

Explanation:

Given mass of block A is  $m_{a} = 3kg$

          mass of cart is  $m_{c} = 2kg$

          coefficient of friction between the cart and the block, $\mu=\frac{2}{5}$.

          initial height, h = 6m

At height h the block possess potential energy, $PE=m_{a} gh$

Sliding down the plane, the block looses potential energy and gains kinetic energy, $KE=\frac{1}{2}m_{a}v^{2}$

The loss in potential energy is equal to the gain in  kinetic energy, i,e.,

                                    $\frac{1}{2}m_{a}v^{2}=m_{a}gh$

                                     $\implies v=\sqrt{2gh}$

                                               $=\sqrt{2\times 10 \times 6}=\sqrt{120}m/s$                

Following the principle of conservation of momentum,

                                   $m_{a} v=(m_{a}+m_{c})v_c}$

                    $\implies 3\times \sqrt{120} =(3+2)v_c}$

                                $\implies v_{c}= \frac{3\times \sqrt{120}}{5} =\frac{6}{5} \sqrt{30} m/s$

The total kinetic energy will be equal to the work done which is equal to force times displacement.

                   $\frac{1}{2}m_{a}v^{2}+\frac{1}{2}(m_{a}+m_{c})v_c}^{2}=\mu mg\times X$

$\frac{1}{2}\times 3\times (\sqrt{120}) ^{2}+\frac{1}{2}(3+2) (\frac{6}{5} \sqrt{120})^{2}=\frac{2}{5} \times 5\times 10 \times X$

$\implies X=117m$

Therefore, the position of the block on the cart after it comes to rest relative to cart is 117m.