A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 1.70 × 1013 m/s2 in a machine. If the proton has an initial speed of 4.10 × 105 m/s and travels 2.30 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

a)9.7 × 10⁻⁵m/s

b)= 6.46 × 10⁻⁶J

Explanation:

a) mass m = 1.67 × 10-27 kg

a =  1.70 × 1013 m/s²

distance change Δs = 2.30cm = 2.3 × 10⁻²m

initial speed v(i) = 4.10 × 10⁵m/s

= (4.10 × 10⁵)² + 2(1.70 × 10¹³)(2.3 × 10⁻²)

b) we calculate the initial kinetic of its motion

we calculate the final kinetic of its motion

Finally we calculate the increase in its kinetic energy

= (7.86 × 10⁻⁶) - (1.4× 10⁻⁶)

= 6.46 × 10⁻⁶J

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