An electron of mass 9.11×10−31 leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 2.65 away. It reaches the grid with a speed of 3.20×106 . The accelerating force is constant. How do i find the acceleration? An electron of mass 9.11×10−31 leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 2.65 away. It reaches the grid with a speed of 3.20×106 . The accelerating force is constant. How do i find the acceleration? @Physics

You'll have to use the Work-Energy theorem. \[\Delta E _{k}=W\]\[\Delta E _{_{k}}=-E _{k0}+E _{k}\] \[\Delta E _{_{k}}=-.5mv _{0}^2+.5mv^2\] \[\Delta E _{k}=0+4.6643\times 10^{-18}\] \[W=F.d\] \[W=(m/a).d\] \[W=(m \Delta v/\Delta t).d\] \[4.6643\times 10^{-18}=[(9.11\times 10^{-31}\times 3.2\times 10^{6})/\Delta t]\times 2.65\] \[\Delta t=1.66\times 10^{-6} s\] \[a=\Delta v/\Delta t\] \[a=3.2\times 10^{6}/1.66\times 10^{-6}\] \[a= 1.93\times 10^{^{12}} m/s^2\]

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