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Question

An object of mass 2 kg is placed at rest on a frictionless horizontal surface. A horizontal force is given by $\bar{F} = (x^{2} - 9) \overset{\land}{ℓ} N$ acts on it along x-axis. Kinetic energy of the object will be maximum at

Moderate

A

$x = + 3 m$
B

$x = - 3 m$
C

$x = + 1.8 m$
D

$x = - 1.8 m$

Solution

$f r o m w o r k e n e r g y t h e o r e m w o r k d o n e = c h a n g e i n K E d (K E) = d W = F . d x \Rightarrow K E = \int F . d x K E = \int (x^{2} - 9) d x = \frac{x^{3}}{3} - 9 x t o f i n d K E i s m a x i m u m o r m i n i m u m, \frac{d}{d x} K E = 0 \Rightarrow (x^{2} - 9) = 0 \Rightarrow x = \pm 3 c a s e 1 : i f x = + 3 \to \frac{d^{2}}{d x^{2}} K E = \frac{d}{d x} (x^{2} - 9) = 2 x = 2 (3) = + 6 i s p o s i t i v e \Rightarrow m i n i m u m c a s e 2 : i f x = - 3 \to \frac{d^{2}}{d x^{2}} K E = 2 (- 3) = - 6 i s n e g a t i v e \Rightarrow m a x i m u m ∴ K E i s m a x i m u m a t x = - 3 m$

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