Question

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive x-axis direction is applied to the block . The force ˆF is given by ˆF=(4−x2)ˆiN . Initial position of the block is x=0. The maximum kinetic energy of block between x=0 to x=2.0 is

Answer 1

hence, maximum kinetic energy of the block between x = 0 to x = 2.0 is 5.33 J.

change in kinetic energy of block , K.E = workdone by block = $\int{(4-x^2)}\,dx$

= 4x - x³/3

kinetic energy will be maximum at a point where dK.E/dx= 0

so, 4 - x² = 0 ⇒x = ±2

at x = +2 , d²K.E/dx² < 0

so, at x = 2 , K.E will be maximum.

then, maximum kinetic energy = 4(2) - (2)³/3

= 8 - 8/3 = (24 - 8)/3 = 16/3 = 5.33 J

hence, maximum kinetic energy of the block between x = 0 to x = 2.0 is 5.33 J

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Answer 2

$\huge\bold\purple{Answer:-}$

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive x-axis direction is applied to the block . The force ˆF is given by ˆF=(4−x2)ˆiN . Initial position of the block is x=0. The maximum kinetic energy of block between x=0 to x=2.0 is.