Example-1
A block of mass m = 1 kg, moving on a horizontal surface with initial speed of 2 m/s enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding
Force F on the block in this range is inversely proportional to x over this range
$F= - \frac {k}{x}$ for 0.1 < x < 2.01 m
= 0 for x < 0.1m and x > 2.01 m
where k = 0.5 J. What is the final kinetic energy and speed of the block as it crosses this patch ?
Solution
From Work Energy theorem
$K_f - K_i = \int_{x_a}^{x_b} fdx$
$K_f - \frac {1}{2} 1 \times 2^2 = \int_{.10}^{2.01} - \frac {k}{x} dx$
or
$K_f = 2 - k ln \frac {2.01}{.1}$
$K_f= .5 J$
Now
$\frac {1}{2}mv_f^2 = .5$
or
$v_f = \sqrt {\frac {2 \times .5}{1}}$
$v_f=1m/s$
Example-2
A Jeep having mass 1000 kg is moving with 30 m/s speed. Suddenly it applies the brakes and skids to rest in certain distance d. The
friction force between the tyres and road is 6000 N. Find the value of d?
Solution
From Work Energy theorem
$K_f - K_i =F.d$
$ 0 - \frac {1}{2} mv^2 = -6000.d$
$\frac {1}{2} 1000 \times 30^2 = 6000d$
d=75 m