In the apparatus shown in the figure (Figure 1) , m1 = 10kg and the coefficients of static and kinetic friction between m1 and the table are 0.69 and 0.32, respectively.
Part A
What mass of m2 will just barely set the system in motion?
W1 = m * g = 10kg * 9.8N/kg = 98 N. = Wt. 0f m1.
Fs = u * mg = 0.69 * 98 = 67.62 N. = Force of static friction.
m2 * g = Fs = 67.62 N.
m2 * 9.8 = 67.62
m2 = 6.9 kg.
Well, in the spirit of comedy, let's give this question a funny twist! So, the mass of m2 that will just barely set the system in motion is... drumroll, please... "zero"!
Yep, you heard it right! If the mass of m2 is zero, then it won't contribute any weight or force to the system, making it easier for m1 to push it and set the whole thing in motion. Just like how a feather can easily be blown away by a gentle breeze!
But hey, don't be disappointed! Remember, this is just a hypothetical scenario to inject some humor into your question. In reality, if you want a more appropriate answer, you'll need to consider the actual weight and forces involved.
To determine the mass of m2 that will just barely set the system in motion, we need to consider the force of static friction between m1 and the table.
The force of static friction can be calculated using the formula: fs = μs * N
where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
The normal force acting on m1 can be calculated as N = m1 * g, where m1 is the mass of m1 and g is the acceleration due to gravity.
Given that m1 = 10 kg and μs = 0.69, we have:
N = m1 * g
N = 10 kg * 9.8 m/s^2
N = 98 N
Substituting the values into the formula for static friction, we have:
fs = μs * N
fs = 0.69 * 98 N
fs = 67.62 N
Since the force of static friction opposes the force applied by m2, the force applied by m2 must be equal to or greater than fs in order to set the system in motion. In other words, the force applied by m2 is equal to fs when the system just begins to move.
The force applied by m2 can be calculated using the formula: F = m2 * a
where F is the force applied by m2 and a is the acceleration.
Since the system is just beginning to move, the acceleration is zero (a = 0). Therefore, the force applied by m2 is also zero.
Setting fs equal to the force applied by m2, we have:
fs = F
67.62 N = m2 * 0
Solving for m2, we get:
m2 = 0 kg
Therefore, the mass of m2 that will just barely set the system in motion is 0 kg.
To find the mass (m2) that will just barely set the system in motion, we need to calculate the maximum force of static friction acting on m1.
First, let's determine the maximum force of static friction (F_static_max) using the equation:
F_static_max = coefficient of static friction * normal force
The normal force (N) can be calculated as the weight (mg) of m1:
N = m1 * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, we substitute the given values into the equation to find F_static_max:
F_static_max = 0.69 * (m1 * g)
Now, to just set the system in motion, the force exerted by m2 must be equal to or slightly greater than F_static_max. The force exerted by m2 can be calculated as:
Force (F) = m2 * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the force of static friction (F_static_max) and the force exerted by m2 (F) are equal when the system just starts to move, we can set up the following equation:
F_static_max = F
Substituting in the values, we have:
0.69 * (m1 * g) = m2 * g
Now, we can cancel out the acceleration due to gravity (g) from both sides of the equation:
0.69 * m1 = m2
Finally, we substitute the given value of m1 (10 kg) into the equation to find the mass of m2:
m2 = 0.69 * 10 kg
Calculating the value:
m2 = 6.9 kg
Therefore, a mass of 6.9 kg for m2 will just barely set the system in motion.