Question

A 3.5 kg particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 2.5 m/s at x = 0 m. What is its velocity at x = 2 m and4 m? (Indicate the direction with the sign of your answer. Assume the particle is initially traveling in the positive x-direction.)

Answer 1

From the work - Energy theorem we have :Workdone= change in k.E(a)at x = 2 mfrom right angled triangle ΔOAB ,ForceF =1/2( base*height)=(1/2)(2 * 10)= 10 Nfrom work energy theorem ,work done = change in kinetic energyFS =(1/2)(m)(v₂²-v₁² )10 *2 =(1/2)(3.5kg)(v₂²-(2.5)²)velocity v₂ = 4.20 m/s (along positive x-axis)(b)atx = 4 mfrom right angled triangle ΔOAB ans ΔBCD ,force F= (1/2)(4*10)+(1/2)(4*-10)= 20 - 20 = 0Nfrom work energy theorem ,work done = change in kinetic energyFS =(1/2)(m)(v₂²-v₁² )0 =(1/2)(3.5kg)(v₂²-(2.5)²)(v₂²-(2.5)²) = 0velocity v₂ =2.5 m/s (along positive x-axis)