Physics: 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?

A particle of mass 2.37 kg is subject to a force

Ответ:

zoe2252

20.01.2022

Physics

$v(1.5)=0.7648\ m/s$

Explanation:

Dynamics

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

$\displaystyle a=\frac{F}{m}$

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

F=2cos1.1t

The variable acceleration is calculated by:

$\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}$

a=0.8439cos1.1t

The instant velocity is the integral of the acceleration:

$\displaystyle v(t)=\int_{t_o}^{t_1}a.dt$

$\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt$

Integrating

$\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}$

$\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)$

$\boxed{v(1.5)=0.7648\ m/s}$

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Ответ:

bryceisawesome110

14.09.2021

Physics

Askydiver falls toward the ground at a constant velocity. the statement best applies to newton's second law of motion. newton's second law of motion refers to the behavior of objects' acceleration is dependent upon the net force acting upon the object and the mass of the object.

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