a = 3.6*10^15 m/s^2
accelerating force = M*a = 6.01*10^-12 newtons
Final velocity = V2
Initial Velocity = V1 = 2.4*10^7 m/s
(a) M*V2^2/2 = M*V1^2/2 + F*X
Solve for V2
(b) KE increase = F*X
A proton (mass m = 1.67 10-27 kg) is being accelerated along a straight line at 3.6 1015 m/s2 in a machine. The proton has an initial speed of 2.4 107 m/s and travels 2.8 cm.
(a) What is its speed?
(b) What is the increase in its kinetic energy?
1 answer
1 answer