A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoidally with time. With the positive x-axis pointed towards the East, the x-component of the force is given as follows:
Fx = F₀cos(ωt), where F₀ = 2 N and ω = 1.1 rad/s.
At t = 0 the particle is at x₀ = 0 and has the x-component of the velocity, vₓ = 0.
What is the x-component of velocity (vₓ) in meters per second at t= 1.5 seconds?

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elcharly64 elcharly64 · Answer 1 · 2020-03-10T11:29:18+01:00

Answer:

[tex]v(1.5)=0.7648\ m/s[/tex]

Explanation:

Dynamics

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

[tex]\displaystyle a=\frac{F}{m}[/tex]

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

[tex]F=2cos1.1t[/tex]

The variable acceleration is calculated by:

[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]

[tex]a=0.8439cos1.1t[/tex]

The instant velocity is the integral of the acceleration:

[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]

[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]

Integrating

[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]

[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]

[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]