First slide

Work done by Diff type of forces

Question

A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is

Moderate

A

45 J
B

30J
C

70J
D

20J

Solution

Work done W = Area under F-x graph with proper sign
W = Area of triangle ABC + Area of rectan gle CDEF + Area of rectangle FGHI + Area of rectangle IJKL
W = $[\frac{1}{2} \times 6 \times 10] + [4 \times (- 5)] + [4 \times (5)] + [2 \times (- 5)]$
= 30-20+20-10
= 20J
According to work energy theorem
$K_{f} - K_{i} = W$
or ${(K_{f})}_{x = 16 m} - {(K_{i})}_{x = 0 m} = W$
${(K_{f})}_{x = 16 m} = {(K_{i})}_{x = 0 m} + W$
= 25 J+20 J (Using (i))
= 45 J

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