In a classic laboratory experiment, a cart of mass m1 on a horizontal air track is attached via a string over an ideal pulley to a hanging cylinder of mass m2. the system is released from rest, and the motion is measured. find the speed of the cart after the cylinder has descended a distance h. (use the following as necessary: m1, m2, h, and g.)

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart and the y-axis in the direction the movement of the cylinder

Forces acting on the cart

W₁: Weight of the cart : In vertical direction  (-y)

N₁ : Normal force :In vertical direction  (+y)

T: tension force:  In horizontal direction (+x)

Newton's second law  to the cart:

∑Fx = m₁*a

T = m₁*a Equation(1)

Forces acting on the cylinder

W₂= m₂*g : Weight of m₂ : In vertical direction  

T : Tension force: In vertical direction

Newton's second law  to the cylinder :

We take as positive the forces that go in the same direction of the movement of  the cylinder, that is, downaward:

∑Fy =  m₂*a

W₂-T =  m₂*a

W₂- m₂*a  = T Equation(2)

Equation(1)= Equation(2) =T

m₁*a=W₂- m₂*a

m₁*a+m₂*a = W₂

a( m₁+m₂) = m₂*g

Speed of the cart after the cylinder has descended a distance H

We apply the kinematic equation to the cart:

vf²=v₀²+2*a*d formula (2)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀=0

d = H

We replace data in the formula (2)

vf₁²=v₀₁²+2*a*H

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explanation:

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