Question

A force on a particle depends on position such that f(x) = (3.00 n/m2)x 2 + (3.50 n/m)x for a particle constrained to move along the x-axis. what work is done by this force on a particle that moves from x = 0.00 m to x = 2.00 m?

Answer 1

M = 0.21x ENJOY MY FRIend

Answer 2

Work done, W = 15 joules

Step-by-step explanation:

It is given that,

The force acting on a particle depends on position such that,

`F(x)=3x^2+3.5x`

Let W is the work done by this force on a particle that moves from x = 0.00 m to x = 2.00 m. The expression for work done is given by :

`W=\int\limits^(x_2)_(x_1) {F.dx}`

`W=\int\limits^(2)_(0) {(3x^2+3.5x).dx}`

`W=(x^3+1.75x^2)|^2_0`

W = 15 Joules

So, the work done by this force on a particle is 15 joules. Hence, this is the required solution.