Physics 34 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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