Given,

Mass of the body, m = 2kg

Force applied, F = 7N

Coefficient of kinetic friction, μ = 0.1

A. Time, t = 10s

Using Newton’s second law of motion,

F=ma

where, a is the acceleration caused by force F on a body of mass m

⇒ a_F=F/m

∴

⇒ a=3.5 m/s²

Frictional force is given by

f=μmg

Where, μ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

∴ f=0.1 × 2kg × 9.8ms^-2

⇒ f=1.96N

Acceleration due to friction is given by

a_f=f/m

∴ (The negative sign is due to the fact that the frictional force acts opposite to the direction of motion)

⇒ a_f=-0.98ms^-2

Total acceleration = acceleration due to applied force + acceleration due to friction

Total acceleration, a=a_F+a_f

⇒ a=3.5ms^-2+(-0.98ms^-2)

∴ a=2.52ms^-2

The displacement of the object is given by Newton’s second equation of motion as

⇒ s= 0 + (1/2)(2.52ms^-2)(10s)²

⇒ s=126 m

Work done by the applied force is given by

W=F∙s

⇒ W=7 N × 126 m

⇒ W=882 J

B. Work done by friction is given by

W=f∙s

⇒ W= (-1.96 N) × (126 m)

⇒ W= -246.96 J

C. Net force on the body, F_t=F+f

∴ F_t=7 N + (-1.96 N) = 5.04 N

Work done by total force is given by

W_t=F_t∙s

⇒ W_t=5.04 N × 126 m

⇒ W_t=635.04 J

D. Using newton’s First equation of motion, the final velocity is calculated as

v=u+at

⇒ v=0+(2.52 ms^-2)(10 s)

⇒ v=25.2 m/s

Change in kinetic energy,

So, ΔK = (1/2)(2kg)(25.2ms^-1)² – (1/2)(2kg)(0)

⇒ ΔK = 635.04 J

This result is in accordance with Work-Energy theorem which states that the change in kinetic energy of an object is equal to the net work done on the object.