Answer :
Answer:
Coefficient of friction = 0.20
Acceleration = 3.21m/s^2
Explanation:
Fnet = μkN, where N = weight of upper block (since the normal force is is holding it up against gravity) = 60kg*(9.8 m/s2) = 588 N.
By Newton's 2nd law, Fnet = ma, so that Fnet = (60 kg)(2 m/s2) = 120 N
Thus, 120 N = μk(588 N) --> μk = 120/588 = 0.20
Horizontal Force - Frictional Force = ma (since there it's hard to draw the force diagram here)
310 - (Friction Constant)*(Normal of 60kg block) = (Mass of 60kg block)*(accelaration)
310 - (0.20)×(60 ×9.8) = 60 × a
310 - 117.6 = 60a
192.4 /60 = a
3.21m/s^2 = acceleration
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 = ?
Answer:
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.
