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A man of mass (m_A) is sitting in a train movingwith velocity v_1 = 3 m/s . He observe that a block of mass m_B is moving towards him with velocity v_2 = 5 m/s. Then the kinetic energy of block w.r.t  man is (Assume m _B= 5 Kg)

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S safeer
Answered Sep 09, 2019

 K.E of the block from the reference frame of the train is 

\frac{1}{2} m (v_1+ v_2) ^2

\\=\frac{1}{2} \times 5 \times [ 3+5] ^2 = 1/2 \times 5 \times 64\\ \\ =5 \times 32 = 160 J

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in the figure. Work done by normal reaction on block in time t is:

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S safeer
Answered Sep 09, 2019

By balancing the forces in vertical direction, we get:

\Rightarrow N-Mg=\frac{Mg}{2}\Rightarrow N=\frac{3mg}{2}

and displacement is given by equation:

s=ut+\frac{1}{2}at^2

where u=0 and a=g/2

\Rightarrow\vec{S}=\frac{1}{2}\vec{\frac{g}{2}}t^{2}

and we know,

W_{N}=\vec{N}.\vec{S}

\Rightarrow W=\frac{3mg^{2}t^2}{8}

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