Masses 19 kg and 9 kg are connected by a light string that passes over a frictionless pulley as shown in the figure ( The figure shows the 19 kg block on the table connected by a string through a pulley which is connected to the 5 kg block hanging off the table). The acceleration due to gravity is 9.8m/s. If the 19 kg mass, initially held at rest on the table, is released and moves 1.2 m in 1.4s, determine the coefficient of kinetic friction between it and the table.

Ok, work done by gravity on the 9 kg block is put into friction, and the final KE of both blocks. One then has to find the final velocity of the blocks.

Energy done by gravity= 9g*1.2
energy going into friction= mu*19g*1.2
energy going into ke= 1/2 (27)vfinal^2

distancewent=1/2 a t^2 solve for a
1.2=1/2 a (1.4^2) solve for a.
then vfinal= at= a*1.4
vfinal=1.2*2/1.96 *1.4 check that
vfinal^2= 2.94 check that.
so you have the pieces.
9g*1.2=19g*mu*1.2+1/2 (27)(2.94 )
and you solve for mu

To determine the coefficient of kinetic friction between the 19 kg mass and the table, we can use the equations of motion and Newton's second law.

1. First, we need to calculate the net force acting on the 19 kg mass. It can be determined using Newton's second law, which states that the net force is equal to the mass times the acceleration:

F_net = m * a

Given that the mass of the 19 kg block is 19 kg and it moves with an acceleration of a, we have:

F_net = 19 kg * a

2. The net force acting on the 19 kg mass is equal to the force of gravity pulling it downwards (m1 * g) minus the force of friction opposing its motion (F_friction):

F_net = m1 * g - F_friction

3. Since the 19 kg mass is released and moves downward, the force of friction can be determined as the product of the coefficient of kinetic friction (μ) and the normal force (N):

F_friction = μ * N

The normal force N is equal to the weight of the 19 kg mass, which is given by:

N = m1 * g

4. Combining equations (2) and (3), we have:

m1 * a = m1 * g - μ * m1 * g

Simplifying the equation, we get:

a = g - μ * g

5. We are given that the 19 kg mass moves 1.2 m in 1.4 s. The acceleration (a) can be calculated using the equation:

a = (vf - vi) / t

Where vf is the final velocity, vi is the initial velocity (which is zero as the mass is initially at rest), and t is the time taken. Substituting the given values, we have:

(0 - 0) = (1.2 m - 0) / 1.4 s
0 = 1.2 m / 1.4 s
0 = 0.857 m/s^2

6. Substituting the value of acceleration (0.857 m/s^2) in equation (4), we can find the coefficient of kinetic friction (μ):

0.857 m/s^2 = 9.8 m/s^2 - μ * 9.8 m/s^2
μ * 9.8 m/s^2 = 9.8 m/s^2 - 0.857 m/s^2
μ * 9.8 m/s^2 = 9.8 m/s^2 - 0.857 m/s^2
μ * 9.8 m/s^2 = 8.943 m/s^2
μ = 8.943 m/s^2 / 9.8 m/s^2
μ = 0.913

Therefore, the coefficient of kinetic friction between the 19 kg mass and the table is approximately 0.913.

To determine the coefficient of kinetic friction between the 19 kg mass and the table, we need to follow these steps:

Step 1: Calculate the acceleration of the system.
Since the mass is initially at rest, the final velocity is zero. We can use the formula of motion to calculate the acceleration (assuming constant acceleration):

v = u + at

where:
v = final velocity (0 m/s)
u = initial velocity (unknown, but 0 m/s since it starts from rest)
a = acceleration (unknown)
t = time (1.4 seconds)

Rearranging the equation to solve for acceleration:

a = (v - u) / t

a = (0 - 0) / 1.4
a = 0 m/s² (acceleration is zero since the object comes to rest at the end)

Step 2: Calculate the net force acting on the 19 kg mass.
Since the acceleration is zero, the net force acting on the mass is also zero. At this point, three forces are acting on the 19 kg mass: the force of gravity (mg), the tension in the string, and the force of kinetic friction (fk).

Since the system is in equilibrium (no acceleration), the force of tension is equal to the force of gravity acting on the 9 kg mass (m2g):

Tension = Fg2
Tension = m2g
Tension = 9 kg * 9.8 m/s²
Tension = 88.2 N

Step 3: Calculate the force of kinetic friction (fk).
The force of kinetic friction (fk) is given by the equation:

fk = μk * Normal force

where:
μk = coefficient of kinetic friction (unknown)
Normal force = m1 * g (the force exerted by the table on the 19 kg mass)

Substituting the values into the equation:

fk = μk * m1 * g

We need to find the normal force, which is equal to the weight of the 19 kg mass:

Normal force = m1 * g
Normal force = 19 kg * 9.8 m/s²
Normal force = 186.2 N

Now we can calculate the force of kinetic friction:

fk = μk * Normal force
fk = μk * 186.2 N

Step 4: Calculate the displacement (s) of the 19 kg mass.
We are given that the 19 kg mass moves 1.2 m in 1.4 seconds. This is the displacement of the mass (s). Speed (v) can be calculated using the formula:

v = s / t
v = 1.2 m / 1.4 s
v ≈ 0.86 m/s

Step 5: Calculate the work done by kinetic friction (Wfk).
The work done by kinetic friction (Wfk) is calculated using the equation:

Wfk = fk * s * cos(180°)

Since the force of kinetic friction and the displacement are in opposite directions, the angle between them is 180 degrees.

Wfk = fk * s * cos(180°)
Wfk = fk * 1.2 m * cos(180°)
Wfk = fk * (-1.2 m)

Step 6: Calculate the work done by the net force (Wnet).
The work done by the net force (Wnet) is calculated using the equation:

Wnet = Tension * s * cos(0°)

Since the force of tension and the displacement are in the same direction, the angle between them is 0 degrees.

Wnet = Tension * s * cos(0°)
Wnet = 88.2 N * 1.2 m * cos(0°)

Step 7: Calculate the friction coefficient (μk).
Since the net force is equal to zero (since the acceleration is zero), the work done by the net force (Wnet) is equal in magnitude but opposite in sign to the work done by kinetic friction (Wfk).

Wnet = -Wfk

Substituting the values, we get:

88.2 N * 1.2 m * cos(0°) = - fk * 1.2 m

Simplifying the equation:

88.2 N = - fk

Comparing this equation with fk = μk * 186.2 N, we can find the coefficient of kinetic friction (μk) to be:

μk = 88.2 N / 186.2 N
μk ≈ 0.473

Therefore, the coefficient of kinetic friction between the 19 kg mass and the table is approximately 0.473.