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At, t = 0 force F→=(-5.00i^+5.00j^+4.00k^)Nbegins to act on a 2.00 kg particle with an initial speed of 4.00 m / s . What is the particle’s speed when its displacement from the initial point is role="math" localid="1657949233150" d→=(2.00i^+2.00j^+7.00k^)m?

Short Answer

Expert verified

Speed of the particle is 6.63 m/s.

Step by step solution

01

Given information

It is given that,

  1. Force isF→=-5.00i^+5.00j^+4.00k^N
  2. Displacement is d→=(2.00i^+2.00j^+7.00k^)m
  3. Mass is given asm = 2.00 kg
  4. Initial velocity is given as.vi=4.00m/s
02

Determining the concept

The problem deals with the work done which is the fundamental concept of physics. Work is the displacement of an object when force is applied on it. First, findthework done bythedot product ofthegiven vectors, and then find the final velocity of the particle.

Formulae are as follow:

W=FdW=12mvf2-12mvi2

Where,

F istheforce, vfviarethefinal and initial velocities, m isthemass, d isthedisplacement and W is the work done.

03

Determining the speed of the particle

Thework done can be found as,

W=F→.d→W=-5.00i^+5.00j^+4.00k^.2.00i^+2.00j^+7.00k^W=28.0J

Using this value, find the velocity,

W=12mvf2-12mvi228.0=12(2.00)vf2-12(2.00)(4.00)2 28.0=vf2-16vf2=28.0+16vf=6.63m/s

Hence, the speed of the particle is vf=6.63m/s.

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