A proton is accelerated at 3•6×10^15ms^2 through a distance of 3•5cm. if the initial velocity is 2•4×10^9ms find the change in kinetic energy at the end of the distance. Taking mass of proton =1•6×10^-27kg
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Answer:
20.16 × 10⁻¹⁴ J
Explanation:
Given :
A proton is accelerated at 3.6×10¹⁵ m/s²,
through a distance of 3.5cm.
If the initial velocity is 2.4×10⁹m/s.
Find the change in kinetic energy at the end of the distance.
(Taking mass of proton =1•6×10^-27kg)
Solution :
We know that,
Work done = Change in Kinetic energy = Force × Displacement = m a × s
⇒ (1.6 × 10⁻²⁷) × (3.6 × 10¹⁵) × (3.5 × 10⁻²)
⇒ (1.6 × 3.6 × 3.5) × (10⁻²⁷⁺¹⁵⁻²)
⇒ (5.76 × 3.5) × (10⁻¹⁴)
⇒ 20.16 × 10⁻¹⁴ J
(or)
Change in kinetic energy =
⇒
⇒
We know that,
v² - u² = 2as,
So,
⇒
⇒ , which is same,.
⇒ (1.6 × 10⁻²⁷) × (3.6 × 10¹⁵) × (3.5 × 10⁻²)
⇒ (1.6 × 3.6 × 3.5) × (10⁻²⁷⁺¹⁵⁻²)
⇒ (5.76 × 3.5) × (10⁻¹⁴)
⇒ 20.16 × 10⁻¹⁴ J
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