Question

A body of weight 1 kN is on the horizontal surface
of a table. This weight is connected to another
body of weight 2 kN by a string passing over a
smooth pulley fixed at the corner of the table. The
coefficient of friction between 1 kN weight and the
table surface is 0.20. If the system is released from
the rest, find the velocity of the 2 IN weight after
it has moved 1.2 m using the work-energy method.​

Answer 1

Answer :

Weight of block A = 2000N

Weight of block B = 1000N

Coefficient of kinetic friction b/w block B and surface = 0.20

We have to find velocity of block B when it covered 1.2m distance$.$

★ First of all we need to find common acceleration of the system.

➠ m₁g - fk = (m₁ + m₂)a

• Mass of A = 2000/10 = 200kg
• Mass of B = 1000/10 = 100kg

➠ (200)(10) - 200 = (200+100)a

➠ a = (2000-200)/300

➠ a = 1800/300

➠ a = 6m/s²

★ As per work energy theorem,

• work = ∆(kinetic energy)

➙ 1/2m₂v² - 1/2m₂u² = F × d

• Initial velocity = zero

➙ 1/2 m₂v² - 0 = m₂a × d

➙ v²/2 = ad

➙ v² = 2ad

➙ v² = 2(6)(1.2)

➙ v = √14.4

➙ v = 3.8 m/s