Answer:
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
[tex]v_f^2 = v_i^2 + 2as[/tex]
= (4.10 × 10⁵)² + 2(1.70 × 10¹³)(2.3 × 10⁻²)
[tex]v_f = \sqrt{9.5 * 10^1^1} \\ = 9.7 * 10^5[/tex]
b) we calculate the initial kinetic of its motion
[tex]K_i = \frac{1}{2} m_pv_f^2\\= \frac{1}{2} * (1.67 * 10^-^2^7)(4.1 * 10^5)^2\\= 1.4 * 10^-^1^6J[/tex]
we calculate the final kinetic of its motion
[tex]K_i = \frac{1}{2} m_pv_f^2\\= \frac{1}{2} * (1.67 * 10^-^2^7)(9.7 * 10^5)^2\\= 7.86 * 10^-^1^6J[/tex]
Finally we calculate the increase in its kinetic energy
= (7.86 × 10⁻⁶) - (1.4× 10⁻⁶)
= 6.46 × 10⁻⁶J
