Verified answer

Given :

Mass of the block = 10g = 0.01kg

Coefficient of kinetic friction = 0.4

Applied force = 50N

To Find :

Work done by each individual force acting on the force over displacement of 5m.

Solution :

★ Work done is measured as the dot product of force and displacement.

• It is a scalar quantity having only magnitude.
• SI unit : Joule (J)
• Dimension formula : [M¹L²T‾²]

Mathematically,

Where θ denotes angle between force and displacement vectors.

A] Work done by external force :

B] Work done by frictional force :

C] Work done by net force :

➠ W = W₁ + W₂

➠ W = 250 + (-0.2)

➠ W = 249.8 N

D] Work done by weight force :

Solution

Net force acting on the block

F - F = ma

50 - ųmg = ma

50 - 0.44 × 10 × 10 = ma

50 - 40 = ma

Net force = 50 - 40 = 10 N

Work done by constant force = F × displacement

50 × 5 = 250 J

Work done by friction :

= -F × 5

= -40 × 5 = -200J

It is negative Since displacement is opposite of the direction of friction

Net work = 250 - 200 = 50J