Proton speed and kinetic energy after acceleration over 3.1 cm

[GalacticSnipes]

Homework Statement

A proton (mass m = 1.67e-27 kg) is being accelerated along a straight line at 2.4e15 m/s2 in a machine. The proton has an initial speed of 2.4e7 m/s and travels 3.1 cm.

(a) What is its speed?

(b) What is the increase in its kinetic energy?

Homework Equations

V^2 = Vo^2+2ad
KE = 0.5m(Vavg)^2
KE = 0.5m(Vf^2-Vi^2)

The Attempt at a Solution

(a) V^2 = 2.4e7 + 2(2.4e15)(.031)
V^2 = 7.248e14
V = 2.69221e7

(b) KE = 0.5(1.67e-27)((2.69221e7)^2-(2.4e7)^2)
KE = 1.24248e-13

 

[gneill]

Your calculations look fine, but you've left off units and specified too many significant figures in the results.

 

[GalacticSnipes]

So for (a) 2.7e7 m/s and for (b) 1.2e-13 J?

 

[gneill]

Yes.

 

[GalacticSnipes]

ok thx