Question

A block of mass m = 1 kg moving on horizontal surface with speed u = 2 m/s enters a rough horizontal patch ranging from x = 0.10 m to x = 2.00 m. If the retarding force f_r on the block in this range is inversely proportional to x over this range i.e.

f_r = `"-k"/x`  0.10 < x < 2.00

= 0 for x < 0.10 and x > 2.00

If k = 0.5 J then the speed of this block as it crosses the patch is (use ℓn 20 = 3)

पर्याय

• 2.65 m/s

• 1 m/s

• 1.5 m/s

• 2 m/s

Answer 1

1 m/s

Explanation:

Use W = ∆K

where W = `int_0.1^2.0`f.dx