| Two blocks of masses \[{{m}_{1}}=1\,\,kg\] and \[{{m}_{2}}=2\,\,kg\] are connected by a non-deformed light spring. They are lying on a rough horizontal surface. The coefficient of friction between the blocks and the surface is 0.4. What minimum constant force F has to be applied in horizontal direction to the block of mass mi in order to shift the other block? \[(g=10m/{{s}^{2}})\] |
 |
A) 8 N
B) 15 N
C) 10 N
D) 25 N
Correct Answer: B
Solution :
| From work-energy theorem; |
| Work done by the all the forces = change in kinetic energy |
| i.e., \[F.x-\mu {{m}_{1}}gx-\frac{1}{2}k{{x}^{2}}=0\] But \[kx=\mu {{m}_{2}}g\] for just shifting \[{{m}_{2}}.\] |
| \[\therefore \,\,\,\,Fx-\mu {{m}_{1}}gx-\frac{1}{2}\mu {{m}_{2}}gx=0\] |
| \[F=\mu \left( {{m}_{1}}+\frac{{{m}_{2}}}{2} \right)g=0.4\left( 1+\frac{2}{2} \right)(10)=8N\] |
You need to login to perform this action.
You will be redirected in
3 sec
