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Question

Physics Question on laws of motion

Two masses m1m_1 = 5 kg and m2m_2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move? (g = 9.8 ms2ms^{-2})

A

0.2 ms2ms^{-2}

B

9.8 ms2ms^{-2}

C

5 ms2ms^{-2}

D

4.8 ms2ms^{-2}

Answer

0.2 ms2ms^{-2}

Explanation

Solution

On releasing, the motion of the sxstem will be according to figure m1gT=m1a...(i)\, \, \, \, \, \, \, \, \, \, \, \, m_1 g - T = m_1 a \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i) and Tm2g=m2a...(ii) \, \, \, \, \, \, \, \, \, \, \, T - m_2g = m_2 a \, \, \, \, \, \, \, \, \, \, \, \, ...(ii) On solving; a=(m1m2m1+m2)g...(iii) \, \, \, \, \, \, \, \, \, \, \, \, \, a =\bigg( \frac{m_1 - m_2}{m_1 + m_2} \bigg)g \, \, \, \, \, ...(iii) Given, m1=5kg,m2=4.8kg,\, \, \, \, m_1 = 5 kg, m_2 = 4.8 kg, g=9.8ms2 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, g = 9.8 ms^{-2} a=(54.85+4.8)×9.8\therefore \, \, \, \, \, \, a =\bigg( \frac{5 - 4.8}{5 + 4.8} \bigg) \times 9.8 0.29.8×9.8=0.2ms2 \, \, \, \, \, \, \, \, \, \, \, \, \frac{0.2}{9.8} \times 9.8 = 0.2 ms^{-2}