Two blocks are arranged at the ends of amassless string as shown in the figure. The system
starts from rest. When the 1.67 kg mass has fallen through 0.395 m, its downward speed is
1.31 m/s.
The acceleration of gravity is 9.8 m/s2.
What is the frictional force between the
3.34 kg mass and the table?
I don't understand the figure, since it isn't shown. Is 3.34 kg on a flat horizontal surface and is 1.67 kg pulling it over to the edge? Does the string go over a pulley? With no pulley, one would expect a lot of friction where the string goes over the edge.
Both masses accelerate at a rate a given by
V^2 = 2 a X
a = V^2/(2X) = 1.658 m/s^2
Write free body diagram equations for both masses, eliminate the spring tension and solve for the friction force
the 3.34 is on a horizontal table and the 1.67 is over the edge. they are attached by a string and I need to know the friction of the horizontal mass (3.34 kg) has on the table with the vertical mass (1.67 kg) hanging off the table.
Why did the 3.34 kg mass go to the table for help? Because it needed some frictional force in its life! But don't worry, I'm here to calculate it for you.
To find the frictional force between the 3.34 kg mass and the table, we need to understand the forces acting on the system. We have the gravitational force pulling down on both masses, and the tension in the string pulling up. Since the 1.67 kg mass is falling, we can assume there is no friction acting on it.
Let's start by finding the net force on the system using Newton's second law: F_net = (m_2 * g) - (m_1 * a)
Where:
- F_net is the net force on the system
- m_1 is the mass of the 1.67 kg block
- a is the acceleration of the 1.67 kg block
Given that the downward speed of the 1.67 kg mass is 1.31 m/s and it has fallen through a distance of 0.395 m, we can find the acceleration using the formula: a = (v_final^2 - v_initial^2) / (2 * d)
Where:
- v_final is the final velocity of the 1.67 kg mass (1.31 m/s)
- v_initial is the initial velocity of the 1.67 kg mass (0 m/s)
- d is the distance the 1.67 kg mass has fallen (0.395 m)
Using this information, we can calculate the acceleration as follows:
a = (1.31^2 - 0^2) / (2 * 0.395)
Now that we know the acceleration, we can calculate the net force on the system:
F_net = (3.34 * 9.8) - (1.67 * a)
And finally, let's solve for the frictional force, which is equal in magnitude to the net force on the 3.34 kg mass:
Frictional Force = F_net
Solving these equations should give you the value of the frictional force between the 3.34 kg mass and the table. I hope this helps, and please remember to support your masses emotionally as well!
To find the frictional force between the 3.34 kg mass and the table, we can use the concept of Newton's second law of motion.
Step 1: Determine the net force acting on the 3.34 kg mass.
Assuming the only forces acting on the mass are the gravitational force and the frictional force, we can write the equation:
Net Force = Weight - Frictional Force
The weight of an object can be calculated using the equation:
Weight = mass * acceleration due to gravity
Given that the mass is 3.34 kg and the acceleration due to gravity is 9.8 m/s^2, we can calculate the weight:
Weight = 3.34 kg * 9.8 m/s^2
Step 2: Calculate the net force acting on the 3.34 kg mass.
We know that the net force is equal to the product of mass and acceleration. Given that the downward speed of the 1.67 kg mass is 1.31 m/s and it falls through 0.395 m:
Net Force = mass * acceleration = mass * (final velocity - initial velocity) / time
Here, initial velocity is 0 m/s, final velocity is 1.31 m/s, and time is the time taken by the 1.67 kg mass to fall 0.395 m. We can calculate time using the equation:
time = (distance) / (velocity)
time = 0.395 m / 1.31 m/s
Step 3: Calculate the frictional force.
To find the frictional force, rearrange the net force equation:
Frictional Force = Weight - Net Force
Substitute the values into the equation and calculate the result to find the frictional force between the 3.34 kg mass and the table.
To solve this you have to know something about the additional friction where the string goes over the edge. Perhaps they want you to assume it is zero. That is why I asked if there is a pulley wheel at the edge.
If there is no string friction,
1.67 g - T = 1.67 a
T - Ff = 3.34 a
1.67 g - Ff = 5.01 a
I already solved for the acceleration a.
Ff = 1.67 g - 5.01 a = 8.06 newtons