Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 1.78 kg mass has fallen through 0.438 m its downward speed is 1.27 m/s. The acceleration of gravity is 9.8 m/s^2. What is the frictional force between the 3.13 kg mass and the table? answer in units of newtons.
To find the frictional force between the 3.13 kg mass and the table, we need to consider the forces acting on the system.
1. Gravity: The force of gravity acts on both masses and can be calculated using the formula F = m * g, where m is the mass and g is the acceleration due to gravity. For the 1.78 kg mass, the gravitational force is F1 = (1.78 kg) * (9.8 m/s^2) = 17.484 N. For the 3.13 kg mass, the gravitational force is F2 = (3.13 kg) * (9.8 m/s^2) = 30.674 N.
2. Tension in the string: The string connecting the two blocks exerts a tension force. Since the system is not accelerating vertically, the tension is constant throughout the string. This tension force acts on both masses in opposite directions.
3. Frictional force: There is friction acting between the 3.13 kg mass and the table. This frictional force opposes the motion of the mass.
Since the system is in equilibrium when the 1.78 kg mass has fallen through 0.438 m with a downward speed of 1.27 m/s, we can equate the forces acting on the masses.
For the 1.78 kg mass:
F1 - T = m1 * a1, where a1 is the vertical acceleration.
For the 3.13 kg mass:
T - F_friction = m2 * a2, where a2 is also the vertical acceleration but with opposite sign.
Since both masses have the same acceleration, we can express the equations in terms of a:
F1 - T = m1 * a
T - F_friction = -m2 * a
Manipulating these equations, we can solve for the frictional force:
F_friction = T + m2 * a
Now, we need to determine the tension in the string and the vertical acceleration.
To find the tension T in the string, we can use the motion of the 1.78 kg mass:
m1 * g - T = m1 * a1
Substituting the values, we get:
(1.78 kg) * (9.8 m/s^2) - T = (1.78 kg) * a1
To find the acceleration a1, we can use kinematic equations. We know that the initial velocity is 0 since the system starts from rest. We also know the final velocity of the 1.78 kg mass is 1.27 m/s when it falls through 0.438 m.
Using the equation v^2 - u^2 = 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can solve for a:
(1.27 m/s)^2 - (0 m/s)^2 = 2 * a1 * (0.438 m)
Simplifying the equation, we find:
a1 = (1.27 m/s)^2 / (2 * 0.438 m) = 1.8537 m/s^2
Substituting the values of m1, g, and a1 into the equation for T, we have:
(1.78 kg) * (9.8 m/s^2) - T = (1.78 kg) * (1.8537 m/s^2)
Simplifying the equation, we find:
T = (1.78 kg) * (9.8 m/s^2) - (1.78 kg) * (1.8537 m/s^2) = 13.1188 N
Now that we have the tension T and the vertical acceleration a, we can substitute them into the equation for the frictional force:
F_friction = (13.1188 N) + (3.13 kg) * (1.8537 m/s^2)
Calculating the equation, we find:
F_friction = 13.1188 N + 5.790301 N = 18.9091 N
Therefore, the frictional force between the 3.13 kg mass and the table is approximately 18.9091 N.
change in potential energy = -1.78 (9.81) (0.438) = -7.65 Joules
increase in kinetic energy = (1/2)(1.78+3.13) (1.27)^2 = 3.96 Joules
Oh my, we lost energy = 7.65-3.96 = 3.69 Joules wasted in friction
so
3.69 = F* distance = F * 0.438
F = 8.43 Newtons