Question

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.)

Answer 1

`v_(f1) =\sqrt{2*((m_(2)*g )/(mi+m2))*H }`

Step-by-step 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

`a= (m_(2)*g )/(m_(1)+m_(2)  )`

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

`a= (m_(2)*g )/(m_(1)+m_(2)  )`

v₀=0

d = H

We replace data in the formula (2)

vf₁²=v₀₁²+2*a*H

`v_(f1) =\sqrt{2*((m_(2)*g )/(mi+m2))*H }`