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A particle of mass m=1 follows the trajectory r=( 3

4

t 3
+t 2
+1)i+ 3
4

(t+4) 3/2
j+2ln(t+1)k Find the particle velocity and the force acting on the particle at all time t, and hence calculate the work done on the particle between t=0 and t=1 a) Velocity and force Calculate the particle's velocity v and the force F acting on the particle. Represent your answers in the form v=
and F=

i+j+k
i+j+k.

Calculate a scalar product of force and velocity: F⋅v= Now calculate the work W=

1 Answer

4 votes

Answer:

Step-by-step explanation:

To find the particle's velocity and the force acting on it, we need to differentiate the given trajectory equation with respect to time (t).

Given trajectory: r = (3t^3 + t^2 + 1)i + (3(t + 4)^(3/2))j + 2ln(t + 1)k

Velocity (v):

The velocity vector (v) is obtained by taking the derivative of the position vector (r) with respect to time (t).

v = dr/dt = (d/dt(3t^3 + t^2 + 1))i + (d/dt(3(t + 4)^(3/2)))j + (d/dt(2ln(t + 1)))k

Differentiating each component:

v = (9t^2 + 2t)i + (3/2)(t + 4)^(1/2)j + (2/(t + 1))k

Therefore, the particle's velocity is v = (9t^2 + 2t)i + (3/2)(t + 4)^(1/2)j + (2/(t + 1))k.

Force (F):

The force acting on the particle is the derivative of momentum with respect to time (F = dp/dt). Since the particle's mass (m) is given as 1, force is equivalent to acceleration (F = ma).

Therefore, we need to find the acceleration vector (a).

Acceleration (a) is obtained by taking the derivative of the velocity vector (v) with respect to time (t).

a = dv/dt = (d/dt(9t^2 + 2t))i + (d/dt((3/2)(t + 4)^(1/2)))j + (d/dt(2/(t + 1)))k

Differentiating each component:

a = (18t + 2)i + (3/4)(t + 4)^(-1/2)j - 2/(t + 1)^2k

Therefore, the force acting on the particle is F = (18t + 2)i + (3/4)(t + 4)^(-1/2)j - 2/(t + 1)^2k.

Scalar Product of Force and Velocity (F ⋅ v):

The scalar product of two vectors is calculated by multiplying their corresponding components and adding them up. For F ⋅ v, we have:

F ⋅ v = (18t + 2)(9t^2 + 2t) + (3/4)(t + 4)^(-1/2)(3/2)(t + 4)^(1/2) + (-2/(t + 1)^2)(2/(t + 1))

Simplifying and combining terms:

F ⋅ v = 162t^3 + 36t^2 + 18t + (9/4) + (-4/(t + 1)^3)

Work (W):

To calculate the work done on the particle between t = 0 and t = 1, we integrate the scalar product F ⋅ v with respect to time (t) over the given interval:

W = ∫[0,1] (162t^3 + 36t^2 + 18t + (9/4) - 4/(t + 1)^3) dt

Integrating each term separately and evaluating the definite integral over [0,1], you can find the numerical value of the work done on the particle between t

User Joao Gavazzi
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