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

Answer:

Work done, W = 15 joules

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.