question_answer

Two identical 5kg blocks are moving with same speed of 2m/s towards each other along a frictionless horizontal suface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by (i), external forces and (ii), internal forces.

Answer:

                As no external force is applied on the system,                 According to work energy principle, total work done = change In K.E. = final K.E. ? initial K.E.                 \[\therefore \] \[{{W}_{ext}}+{{W}_{\operatorname{int}}}=0-\left( \frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}m{{\upsilon }^{2}} \right)=-m{{\upsilon }^{2}}\]     or \[0+{{W}_{\operatorname{int}}}=-5{{\left( 2 \right)}^{2}}=-20J\], i.e., \[{{W}_{\operatorname{int}}}=-20J\] Negative sign implies that internal force of action and reaction act on the two blocks in a direction opposite to their motion.