A particle of mass m, initially at rest, is acted on by a force F = F 0 { 1 − ( T 2 t − T ) 2 } during the interval 0 ≤ t ≤ T . The velocity of the particle at the end of the interval is :

v = m 1 ∫ 0 T F d t = m F 0 ∫ 0 T [ 1 − ( T 2 t − T ) 2 ] d t . Put u = T 2 t − T , d t = 2 T d u , limits − 1 to 1 : ∫ ( 1 − u 2 ) d u = 2 − 3 2 = 3 4 . v = m F 0 ⋅ 2 T ⋅ 3 4 = 3 m 2 F 0 T .

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PhysicsNLMSimple Application of NLMMedium2 minQB

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A particle of mass m, initially at rest, is acted on by a force $F = F_{0} {1 - (\frac{2 t - T}{T})^{2}}$ during the interval $0 \leq t \leq T$ . The velocity of the particle at the end of the interval is :

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Solution:

$v = \frac{1}{m} \int_{0}^{T} F d t = \frac{F _{0}}{m} \int_{0}^{T} [1 - (\frac{2 t - T}{T})^{2}] d t$ .

Put $u = \frac{2 t - T}{T}$ , $d t = \frac{T}{2} d u$ , limits $- 1$ to $1$ : $\int (1 - u^{2}) d u = 2 - \frac{2}{3} = \frac{4}{3}$ .

$v = \frac{F _{0}}{m} \cdot \frac{T}{2} \cdot \frac{4}{3} = \frac{2 F _{0} T}{3 m}$ .

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Simple Application of NLM

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