The force on a particle is directed along an x axis and given by F = F0(x/x0 - 1) where x is in meters and F is in Newtons. If F0 = 1.2 N and x0 = 3.3 m, find the work done by the force in moving the particle from x = 0 to x = 2x0 m.

0 j

Explanation:

Work done from point p1 to point p2 is given by

Work = ∫ F.dr

Here we are given

F= F₀ (x/x₀ -1)

F₀=1.2 N

x₀=3.3 m

dr= dx (since distance changes along x-axis only)

We will calculate above integral between given points x=0 and x= 2x₀

Putting value of F  , x₀ and  dr

==> Work = ∫ F₀ (x/x₀ -1)

=∫ 1.2 (x/3.3 - 1) dx

= 1.2 /3.3

=∫(x-3.3)dx

=0.363636∫(x-3.3)dx

=0.363636 [ x²/2 - 3.3x]

we are given lower point value x= 0

upper point value x= 2 x₀ = 2× 3.3 = 6.6

putting lower and upper point values of x

= 0.363636 [ {0²/2-3.3(0)} - { (6.6)² /2 - 3.3 × 6.6}]

=0.363636 [ {0}- {43.56 /2 - 21.78}]

=0.363636 [ -{0}]

=0.363636 [0]

=0 j

i-e work is zero joules or no work !

are there options? cause you ain’t. real specific

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray