Question

A force acts on a 3g particle in such a way that the position of the particle as a function of time is given by x=3t-4t 2 +t 3 where x is in meters and t is in seconds. The work done during the first 4 sec.

Answer 1

When t=0, x will be 0 and thus it means no work has been done so the initial velocity is also 0
when t=4
x= 3t -4t² +t³
x= 3(4) - 4(4)² + (4)³
x= 12m

so final velocity = x/t = 12/4 = 3

acceleration = a= v-u/t
= 3-0/4
3/4

force = F= ma
= 3 x 10⁻³ x 3/4
= 9/4 x 10⁻³
= 2.25 x 10⁻³

Answer 2

Answer:

0.528 J

Explanation:

we have,

mass = 3g = 0.003kg

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

now

v= dx/dt = 3-8t+3t^2

dx = (3-2t+3t^2)dt

a= dv/dt = -8+6t

now

dw = fdx

dw= (ma)dx

dw = (0.003) (-8+6t) (3-8t+3t^2)dt

dw = (0.003) ∫{(-8+6t) (3-8t+3t^2)}dt

dw = 0.003) [{(-8+6t) (3-8t+3t^2)}]^4to 0

dw = 0.003 × 176 = 0.528