JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 15)
Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in the figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be :
_26th_February_Evening_Shift_en_15_1.png)
$${{MF} \over {F + Ma}}$$
$${{F + Ma} \over M}$$
$${{Ma - F} \over M}$$
$${{F - Ma} \over M}$$
Explanation
Let spring force acting on the both block is Fs.
F.B.D. of Block A
_26th_February_Evening_Shift_en_15_4.png)
$$ \therefore $$ Fs = Ma ..... (i)
F.B.D. of Block B,
_26th_February_Evening_Shift_en_15_5.png)
$$ \therefore $$ Acceleration of Block B,
$${a_1} = {{{F_{net}}} \over M} = {{F - {F_s}} \over M} = {{F - Ma} \over M}$$
F.B.D. of Block A
$$ \therefore $$ Fs = Ma ..... (i)
F.B.D. of Block B,
$$ \therefore $$ Acceleration of Block B,
$${a_1} = {{{F_{net}}} \over M} = {{F - {F_s}} \over M} = {{F - Ma} \over M}$$
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