a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to as second block (m2), which hangs vertically.
Ignore friction and the masses of the pulley and cord.
If m1 = 13.0 kg and m2 = 6.0 kg determine the accleeration of each block
for this I got about 3.09 s^-2 m
If initally m1 is at rest 1.250 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?
for this I got about .90 s
ok this part I need help and possibly the other ones that I just did...
If m2 = 1.0 kg how large must m1 be if the acceleration of the system is to be kept at 100^-1 g
I imediately threw up the flag here because acording to our book the text says when g is not given use the magnitude of 9.80 so I took 100^-1 of that magnitude and got exactly 99 kg
can someone please confirm these answers?
Thank you!
I'm confused, there is no explanation here and I REALLY need help on it!
The answers are correct. In the last part, your answer does not depend on your choice for g. If you leave it undetermined it will factor out of your equations.
I think you should practice a bit with more complicated problems. I think that you get bogged down a bit too much in elementary math and that prevents you from mastering the actual physics.
A good problem would e.g. be to take the cord to be elastic such that the tension in the cord is given by
c(L-L0), where L is the length of the cord.
Write down the differential equations for the position of the two masses in this case. What are the initial conditions if mass 1 is held in place and mass 2 hangs and then we suddenly release mass 1 at t = 0?
Actually, this problem would be a bit too complicated as the tension in the string would not be uniform.
To determine the acceleration of each block in the given system, we can apply Newton's second law of motion. The force acting on each block is given by the tension in the rope.
For block m1:
Tension = net force = mass x acceleration
Tension = m1 x a
For block m2:
Tension = net force = mass x acceleration
Tension = m2 x g, where g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the two blocks are connected by the same cord, the tension in the rope is the same for both blocks. Therefore, we can equate the two tension equations:
m1 x a = m2 x g
Now let's substitute the given values:
m1 = 13.0 kg
m2 = 6.0 kg
g = 9.8 m/s^2
13.0 kg x a = 6.0 kg x 9.8 m/s^2
Simplifying the equation:
13.0a = 58.8
Dividing both sides by 13.0:
a ≈ 4.52 m/s^2
Thus, the acceleration of each block is approximately 4.52 m/s^2.
Now, let's move on to the second part of the question.
To find the time it takes for block m1 to reach the edge of the table, we need to determine the distance it needs to travel. The distance can be calculated using the equation:
Distance = initial position + 0.5 x acceleration x time^2
Given:
Initial position (x0) = 1.250 m
Acceleration (a) = 4.52 m/s^2
We know the distance is equal to the edge of the table, so Distance = x
Substituting the values into the equation:
x = 1.250 + 0.5 x 4.52 x t^2
Now, assuming m1 starts from rest, its initial velocity (v0) is 0 m/s. We can use the equation for velocity:
v = v0 + a x t
0 = 0 + 4.52 x t
Simplifying:
4.52 t = 0
Since t cannot be zero (as the block needs time to reach the edge), we can ignore this solution. Therefore, we know t must be greater than zero.
Now let's substitute the time (t) we found into the distance equation:
x = 1.250 + 0.5 x 4.52 x t^2
Since we want to find the time it takes to reach the edge of the table (x), we set the distance equal to the edge of the table (let's call it L):
L = 1.250 + 0.5 x 4.52 x t^2
To solve for t, we rearrange the equation:
t^2 = (2(L - 1.250))/(4.52)
Taking the square root of both sides:
t = sqrt((2(L - 1.250))/(4.52))
Let's say the edge of the table (L) is 1.5 meters. Substituting this value into the equation:
t = sqrt((2(1.5 - 1.250))/(4.52))
t ≈ sqrt(0.333/4.52)
t ≈ sqrt(0.0736)
t ≈ 0.271 seconds
Therefore, it takes approximately 0.271 seconds for block m1 to reach the edge of the table if the system is allowed to move freely.
Now let's move on to the third part of the question.
If m2 = 1.0 kg and the acceleration of the system (a) is to be 100^-1 g, we need to find the value of m1.
We are given that a = 100^-1 g, where g = 9.8 m/s^2. Thus, a = (100^-1) x 9.8 m/s^2 = 0.098 m/s^2.
m1 x a = m2 x g
m1 x 0.098 = 1.0 x 9.8
Dividing both sides by 0.098:
m1 = (1.0 x 9.8) / 0.098
m1 = 9.8 / 0.098
m1 ≈ 100 kg
Therefore, for the acceleration of the system to be kept at 100^-1 g, m1 must be approximately 100 kg.
Please note that these calculations assume ideal conditions, ignoring factors such as the mass of the pulley and cord, as stated in the problem.