Question

A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where Fx is in Newtons and x is in meters. Find the work done by this force on the particle as the particle moves from x = 0 m to x = 1.9 m. Answer in units of J.

Answer 1

The work done by the force on the particle is 29.85 J.

Step-by-step explanation:

The work is given by:

`W = ^{x_(2)}_{x_(1)}\int F_(x) dx`

Where:

x₁: is the lower limit = 0 m

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N

`W = ^(1.9)_(0)\int (4.5 + 13.7x - 1.5x^(2)) dx`

`W = 4.5x|^(1.9)_(0) + (13.7)/(2)x^(2)|^(1.9)_(0) - (1.5)/(3)x^(3)|^(1.9)_(0)`

`W = 4.5N(1.9 m) + (13.7N)/(2)(1.9 m)^(2) - (1.5N)/(3)(1.9 m)^(3)`

`W = 4.5N(1.9 m) + (13.7N)/(2)(1.9 m)^(2) - (1.5N)/(3)(1.9 m)^(3)`

`W = 29.85 J`

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!