A 3.0-kg particle starts from rest at x = 0.050 m and movesalong the x axis unde

Question

A 3.0-kg particle starts from rest at x = 0.050 m and movesalong the x axis under the influence of a single forceFx = 6.0 + 4.0x - 3.0x2, where Fxis in newtons and x is in meters. (a) Find the work done by the force as the particle moves fromx = 0.050 m to x = 3.0 m. (b) Find the power delivered to the particle as it passesthrough the point x = 3.0 m. A 3.0-kg particle starts from rest at x = 0.050 m and movesalong the x axis under the influence of a single forceFx = 6.0 + 4.0x - 3.0x2, where Fxis in newtons and x is in meters. (a) Find the work done by the force as the particle moves fromx = 0.050 m to x = 3.0 m. (b) Find the power delivered to the particle as it passesthrough the point x = 3.0 m.

Explanation / Answer

A 3.0-kgparticle starts from rest at x = 0.050 m and moves along
the x axis under the influence of a single force Fx =6.0 + 4.0x - 3.0x2,
where Fx is in newtons and x is inmeters.

(a) Find the work done by the force as theparticle moves from x = 0.050 m to x = 3.0m.

     {Work Done By F} = W =

= 0.053.0Fx dx =

         = 0.053.0 (6.0+ 4.0*x - 3.0*x^2)dx

         = [ 6.0*x + 2.0*x^2 - x^3 ]0.053.0  

         = (6.0*(3.0) + 2.0*(3.0)^2 -(3.0)^3) - (6.0*(0.05) + 2.0*(0.05)^2 -(0.05)^3)

          = 8.695125

         = 8.695 Joules


(b)Find the power deliveredto the particle as it passes through the point x = 3.0m.

     Since {P =F*v}, we must first determine object's velocity
     at {x = 3.0 m}. Because objectbegins at rest, we have:

     {Kinetic Energy @ x = 3.0 m} = {Work Done By F} = W
        ---->    (1/2)*m*v^2 = W
        ---->    v = Sqrt[ 2*W/m ]
        ---->    v = Sqrt[ 2*(8.695125 J)/(3.0 kg) ]
        ---->    v = 2.407644 m/sec
        ---->    {Velocity @ x = 3.0 m} = v = 2.407644 m/sec

     {Power DeliveredBy F @ x = 3.0 m} = P = F*v =
          = {Force @ x = 3.0 m}*{Velocity @ x =3.0 m}
        = (6.0 +4.0*(3.0) - 3.0*(3.0)^2)*(2.407644)
        = -21.67 Watts
                   (If not correct, try +21.67Watts)
              
.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.