Question

A particle of mass m=5x10^-27 kg has an initial speed of 2x105 m/s. The particle travels in a straight line and its speed increases to 4x10^5 m/s on a distance of 10 cm. Assuming that the acceleration is constant, the

force exerted on the particle is:
3x10^-15 N
3x10^-20 N
8.65x10^-23N
1.73x10^-22 N

Answer 1

Step-by-step explanation:

Given:

m = 5·10⁻²⁷ kg

V₁ = 2·10⁵ m/s

V₂ = 4·10⁵ m/s

D = 10cm = 0.10m

_______________

a - ?

The change in kinetic energy is equal to the work:

Ek₂ - Ek₁ = F·D

m·V₂² / 2 - m·V₁² / 2 = F·D

(m/2)·(V₂² - V₁²) = F·D

Force:

F = (m/2)·(V₂² - V₁²) / D

F = m·(V₂² - V₁²) / (2·D)

F = (5·10⁻²⁷)·( (4·10⁵)² - (2·10⁵))/ (2·0.10) ≈ 3·10⁻¹⁴ N