First slide

Numericals based on work energy theorem

Question

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

Easy

A

0, 0
B

0, 20 J
C

0, -20 J
D

20 J, -20 J

Solution

$F_{ext} = 0, W_{ext} = \vec{F_{ext}} . \vec{s} = 0$
By work energy theorem: W = $K_{F} - K_{I}$
$W_{ext} + W_{int .} = 0 - (\frac{1}{2} {mv}^{2} + \frac{1}{2} {mv}^{2})$
$W_{int} = - {mv}^{2} = - 5 \times 2^{2} = - 20 J$

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