1.Three masses, m1, m2, and m3, are lying on a plane without friction and are connected by a string. A force F is applied to the mass m1 so that the whole system accelerates along a straight line. What is the tension of the string T between the masses m1 and m2? What is the acceleration of the system?

2.A cylinder rolls without slipping on a horizontal plane and then up an incline. The linear speed of the cylinder at the beginning is v. What maximal height h will the cylinder reach?

1. the string is accelerating m2 and m3, so tension= (m2+m3)a

but a= F/(m1+m2+m3)

2. Figure the total energy of the rolling cyclinder: KE+ RE

1/2 mv^2 + 1/2 I w^2
1/2 mv^2+ 1/2 m r^2*v^2/r^2

so combine all that, then set it equal to mgh, and solve for h, the height it would up.

1. To determine the tension of the string and the acceleration of the system, we need to apply Newton's second law of motion.

First, let's consider the tension in the string between masses m1 and m2. The tension in the string will be the same throughout, as the string is assumed to be massless and without friction.

The forces acting on mass m1 are the applied force F and the tension T. The forces acting on mass m2 are the tension T and the force of gravity mg.

Applying Newton's second law for mass m1:
F - T = m1 * a

Applying Newton's second law for mass m2:
T - mg = m2 * a

Since the overall system accelerates along a straight line, the acceleration will be the same for all masses. Hence, we can set m1 * a = m2 * a.

Now, solving the above equations simultaneously:
F - T = m1 * a
T - mg = m2 * a
m1 * a = m2 * a

We can cancel out the 'a' term on both sides of the equations, resulting in:
F - T = T - mg

Simplifying further:
F = 2T - mg

Rearranging the equation, we can solve for T:
T = (F + mg) / 2

So, the tension in the string between masses m1 and m2 is T = (F + mg) / 2.

To find the acceleration of the system, we can substitute this value of T back into one of the equations:
F - ((F + mg) / 2) = m1 * a

Solving for 'a', we get:
a = (2F - mg) / (2 * m1)

Therefore, the acceleration of the system is a = (2F - mg) / (2 * m1).

2. To determine the maximal height h that the cylinder will reach, we can apply the principle of conservation of energy.

Since the cylinder rolls without slipping, its kinetic energy is given by both its translational and rotational kinetic energy.

At the beginning, the linear speed v of the cylinder can be related to its angular velocity ω using the relationship v = R * ω, where R is the radius of the cylinder.

The initial kinetic energy of the cylinder is given by:
KE_initial = (1/2) * (1/2) * I * ω^2 + (1/2) * m * v^2
where I is the moment of inertia of the cylinder and m is its mass.

As the cylinder rolls up the incline, its linear speed decreases, reaching zero at the maximal height h. At this point, only potential energy is present.

The final potential energy of the cylinder is given by:
PE_final = m * g * h
where g is the acceleration due to gravity.

According to the conservation of energy principle, the initial kinetic energy is equal to the final potential energy:
KE_initial = PE_final

Substituting the expressions for KE_initial and PE_final, we get:
(1/2) * (1/2) * I * ω^2 + (1/2) * m * v^2 = m * g * h

Simplifying the equation:
(1/4) * I * ω^2 + (1/2) * m * v^2 = m * g * h

We also know the relationship between linear velocity and angular velocity: v = R * ω. Substituting this into the equation, we have:
(1/4) * I * (v / R)^2 + (1/2) * m * v^2 = m * g * h

Now, solve for h:
h = (1/4) * I * (v / R)^2 / (m * g) + (1/2) * (v / R)^2

Therefore, the maximal height h that the cylinder will reach is given by h = (1/4) * I * (v / R)^2 / (m * g) + (1/2) * (v / R)^2.