Question

A force F = (2+ 3x) N acts on a particle in positive x
direction. Work done by this force in displacing the
block from x= 0 to x = 2 m is
8 J
10 J
16 J
12 J​

Answer 1

Answer:

ANS: 10J

integration f(x) dx= 2x+(3x^2)/2

apply limits 0 to 2

w= 4+12/2 - 0

w= 4+6= 10 joules

IF U SUBSTITUTE ANY NUMBER IN FUNCTION IT GIVES APPLIED FORCE ONLY NOT WORK DONE

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Answer 2

The correct option regarding the work done by the force is 10J.

Given:

A force F = (2+ 3x) N acts on a particle in a positive x direction.

To Find:

The Work done by the force.

Solution:

To find the work done by the force we will follow the following steps:

As we know,

Work done is the amount of force that is required to move an object to a certain distance or some angle with the force.

The formula for finding the work done = force × displacement

Or,

$work \: done = ∫f.dx$

According to the question:

F = ( 2 + 3x ) N

Block is displaced from 0 to 2 m

Now,

Putting and integrating values we get,

$work \: done = ∫( 2 + 3x )dx = 2x + \frac{ {3x}^{2} }{2}$

Now,

The limit is from 0 to 2 so, integration is solved by putting values of the upper limit of x and the lower limit of x and then subtracting the upper limit integration values from the lower limit integration result.

Here, the upper limit is 2 and the lower limit is 0.

Now,

$2 \times 2+ \frac{ {3(2)}^{2} }{2} - 2 \times 0 + \frac{ {3 \times 0}^{2} }{2} = 4 + 6 - 0 + 0 = 10 \: joule$

Henceforth, the correct option regarding the work done by the force is 10J.

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