CBSE-XI-Physics
05: Newton's Laws of Motion
- #27In figure (5-E11), m1 = 5 kg, m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.
Figure
Ans : Let the acceleration of the blocks be a.
The free-body diagrams for both the blocks are shown below:
From the free-body diagram,
m1a = m1g + F - T ...(i)
Again, from the free-body diagram,
m2a = T - m2g - F ...(ii)
Adding equations (i) and (ii), we have:
`` a=g\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}``
`` ``
`` \Rightarrow a=\frac{3g}{7}=\frac{29.4}{7}``
`` =4.2\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2}``
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,
m1a = F + m1g
5a = 1 + 5g
`` \Rightarrow a=\frac{5g+1}{5}``
`` =g+0.2\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2}``
Page No 81: