Question

A single force acts on a 2.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t â 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 3.0 s.

Answer 1

Answer:2907 J

Step-by-step explanation:

Given

mass of Particle(m)=2 kg

position of particle is given

`x=3t+4t^2+t^3`

thus
`dx=\left ( 3+8t+3t^2\right )dt`

acceleration of particle is given by

`a=\frac{\mathrm{d^2} x}{\mathrm{d} t^2}=8+6t`

Force on particle

`F=ma=2* (8+6t)=4(4+3t)`

`\int dW=\int_(0)^(3)F.dx=\int_(0)^(3)4\left ( 3t+4\right )\left ( 3+8t+3t^2\right )dt`

`W=(36)/(3)\left ( 3\right )^3+(41)/(2)* 9+(9)/(4)\left ( 3\right )^4+12* 3`

`W=4* 726.75=2907 J`