A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 310 N is applied. The 100 kg block sits on a horizontal frictionless surface, but there is friction between the two blocks. find coefficient of friction and Find the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given Information:

Mass of block 1 = m₁ = 60 kg

Mass of block 2 = m₂ = 100 kg

Acceleration of block 1 = α₁ = 2 m/s²

Horizontal force applied to block 1 = F₁ = 310 N

Required Information:

a) Coefficient of friction = μk = ?

b)  Acceleration of block 2 = α₂ = ?

a) Coefficient of friction = 0.323

b)  Acceleration of block 2 = 1.9 m/s²

Explanation:

a) Please refer to the attached diagram, The forces acting on block 1 are; horizontal applied force to the right (F₁), friction force (Fk) acting opposite to applied horizontal force, normal force acting upward (FN₁) and weight of the block acting downwards (W₁).

The sum of forces along x-axis is given by

∑Fx = m₁α₁

F₁ - Fk = m₁α₁

Where Fk = μk*FN₁

μk is the coefficient of friction

F₁ - μk*FN₁ = m₁α₁  eq. 1

The sum of forces along y-axis is given by

∑Fy = 0

FN₁ - W₁ = 0

We know that W₁ = m₁g

FN₁ - m₁g = 0

FN₁ = m₁g  (substitute in eq. 1)

F₁ - μk*FN₁ = m₁α₁

F₁ - μk*(m₁g) = m₁α₁

F₁ - μkm₁g = m₁α₁

- μkm₁g = m₁α₁ - F₁

- μk = (m₁α₁ - F₁)/m₁g

μk = - (m₁α₁ - F₁)/m₁g

μk = - (60*2 - 310)/60*9.8

μk = - (-190/588)

μk = 0.323

b) Please refer to the attached diagram, The forces acting on block 2 are; normal force acting upward (FN₂) and normal force acting downward due to block 1 (FN₁), weight of the block 2 acting downwards (W₂) and frictional force (Fk) to the right and apposite to frictional force acting on block 1.

The sum of forces along x-axis is given by

∑Fx = m₂α₂

Fk = m₂α₂

α₂ = Fk/m₂

Since we know from part (a) Fk = μk*FN₁ = μk*m₁g

α₂ = μk*m₁g/m₂

α₂ = (0.323*60*9.8)/100

α₂ = 189.92/100

α₂ = 1.9 m/s²

Therefore, the acceleration of 100 kg block is 1.9 m/s² during the time that the 60 kg block remains in contact.

i believe the answer is electric potential.

explanation:

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle