Mass of the body, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, µ = 0.1
Initial velocities, u = 0
Time, t = 10 s

The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:

$\mathrm{a}\text{'} = \frac{\mathrm{F}}{\mathrm{m}} = \frac{7}{2} = 3.5 \mathrm{m}/{\mathrm{s}}^2$

Frictional force is given as:

f = µmg
= 0.1 × 2 × 9.8 = – 1.96 N

The acceleration produced by the frictional force:

$\mathrm{a}\text{'} = -\frac{1.96}{2} = -0.98 \mathrm{m}/{\mathrm{s}}^2$

Total acceleration of the body:

$\mathrm{a} = \mathrm{a}\text{'} + {\mathrm{a}}^{\mathrm{n}}$
$= 3.5 + \left(-0.98\right) = 2.25 \mathrm{m}/{\mathrm{s}}^2$

The distance traveled by the body is given by the equation of motion:

$\mathrm{s} = \mathrm{ut} + \frac{1}{2} {\mathrm{at}}^2$
$= 0 + \frac{1}{2} \times 2.52 \times {\left(10\right)}^2 = 126 \mathrm{m}$

(a) Work done by the applied force, Wa = F × s = 7 × 126 = 882 J

(b) Work done by the frictional force, Wf = F × s = –1.96 × 126 = –247 J

Net force = 7 + (–1.96) = 5.04 N

(c) Work done by the net force, Wnet= 5.04 ×126 = 635 J

(d) From the first equation of motion, final velocity can be calculated as:

v = u + at
   = 0 + 2.52 × 10 = 25.2 m/s

$\mathrm{Change} \mathrm{in} \mathrm{kinetic} \mathrm{energy} = \frac{1}{2} {\mathrm{mv}}^2 - \frac{1}{2} {\mathrm{mu}}^2$
$= \frac{1}{2} \times 2\left({\mathrm{v}}^2 - {\mathrm{u}}^2\right) = {\left(25.2\right)}^2 - 0^2 = 635 \mathrm{J}$