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.

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

Explanation:

The work is given by:  

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

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

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