Question

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of X-axis is applied to the block. The force is given by `vecF = (4 - x^2) hati` N, where x is in metre and the initial position of the block is x = 0. The maximum kinetic energy of the block between x = 0 and x = 2.0 m is

विकल्प

• 2.33 J

• 3.33 J

• 5.33 J

• 6.67 J

Answer 1

5.33 J

Explanation:

From work-energy theorem, kinetic energy of block at x = 0 to x is;

K = `int_0^x (4 - x)^2 dx`

∴ K = `4x - x^3/3`

For K to be maximum, `(dK)/(dx)` = 0

or 4 – x² = 0

or x = ± 2 m

At x = + 2m, `(d^2K)/(dx^2)` is negative.

kinetic energy (K) is at its highest point.

∴ K_max = `(4)(2) - (2)^3/3 = 16/3` J = 5.33 J