Question

a position dependent force F acting on a particle and its force position curve is shown in the figure. work done on the particle when its displacement 0to 5 mt is

Answer 1

Work done in 3secinds

$1 \div 2 \times 10 \times (3 + 1) \\ + 20 \: joules$
Work done in 3-5 seconds
-10J
-5

Total work done 20-10-5

So 5 joules is answer

Answer 2

Answer:

The correct answer will be -

5J work is done when its displacement is 0 to 5 m.

Explanation:

Given in the problem is the force-position graph. The area under the graph will give us the work done.

From position 0m to 3m -

We can see this is a trapezium so the area/ work done will be -

$W_1 = \frac{1}{2} \times 10\times (1+3) = 20\ J$

From position 3m to 4m -

we can see it is a triangle so the work done will be -

$W_2 = \frac{1}{2}\times 1\times -10 = -5\ J$

From position 4m to 5m -

We can see it is a rectangle so the work done will be -

$W_3 = -10\times 1 = -10\ J$

So total work done = $W_1 +W_2+W_3$ = 20 J - 5 J - 10 J = 5 J

So 5J work is done when its displacement is 0 to 5 m.

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