Two masses of 200 grams and 400 grams are attached respectively at the end of the cord that passes through around a negligibly weight and frictionless pulley. a.) Find the acceleration of the system, b.) tension in each cord, c.) the distance and velocity of the masses when released from rest at the end of 10 seconds.
well, sum forces.
.200g-.400g=mass(acceleration)
I took the positive direction as the direction the 200grams is headed.
-.200*9.8=(600)a
a= ....
For tension
tension: on the 200 side:
tension=.2*9.8-
.2(a)above
on the 400 side
tension=.4*9.8+.4a
c.
distance=1/2 a t^2
a.Find the acceleration of the system
F^1=m*a
F^2= m*g
F^1= T-F^2
400a= T-1960
F^1= F^2-T
200 a= 3920-T
200a+1960=3920-400a
a= 49/15=3.266=3.27
acceleration=3.27
b.The tension in each cord
T=200*3.27+1960
Tension=2614
T=3920-400*3.27
Tension=2612
c.The distance and velocity of the masses when released from rest at the end of 10 seconds
S=ut+1/2 at^2
S=1/2*3.27*10^2
Distance (S)=163.5m
v^2=u^2+2as
v=√(0+2*3.27*163.5)
Velocity (v)=32.7m/s^2
To solve this problem, we can use the principles of Newton's Laws of Motion.
a.) Find the acceleration of the system:
According to Newton's Second Law, the acceleration of an object is given by the net force acting on it divided by its mass. The net force is determined by the difference between the two tensions in the cord. Let's consider the smaller mass (200 grams) as Mass A and the larger mass (400 grams) as Mass B.
The force acting on Mass A is its weight (mg), directed downward. The force acting on Mass B is also its weight (mg), but directed upward.
Since the pulley is weightless and frictionless, the tension in the cord on either side of the pulley is the same. Let's call this tension T.
So, the net force on Mass A is the force exerted by T minus its weight, and the net force on Mass B is its weight minus the force exerted by T.
Using Newton's Second Law:
For Mass A: T - mg = ma
For Mass B: mg - T = mb
Where m is the mass and a is the acceleration.
Now let's substitute the values given:
m = 200 grams = 0.2 kg (Mass A)
m = 400 grams = 0.4 kg (Mass B)
g = acceleration due to gravity = 9.8 m/s^2
Solving the equations simultaneously will give us the acceleration of the system.
T - 0.2*9.8 = 0.2a (equation 1)
0.4*9.8 - T = 0.4a (equation 2)
Now, solve these equations to find the value of 'a.'
b.) Find the tension in each cord:
Once you have the acceleration, substitute it back into either equation 1 or 2 to find the value of T.
c.) Find the distance and velocity of the masses after 10 seconds:
To find the distance traveled and velocity of the masses after 10 seconds, we'll need to use the equations of motion. Since the system starts from rest, the initial velocity (u) is zero.
For Mass A:
Distance (S) = ut + (1/2)at^2
Velocity (v) = u + at
For Mass B:
Distance (S) = ut - (1/2)at^2
Velocity (v) = u - at
Now, substitute the values of time (t = 10s) and acceleration (a) into these equations to find the distance and velocity of the masses.
By following these steps and utilizing Newton's Laws of Motion and the equations of motion, you should be able to solve the given problem.
Tension in the cord = T
a is up for mass 1, down for mass 2
mass 1
T - .2*9.81 = .2 * a
mass 2
.4 * 9.81 - T = .4 a
solve
2 T - .4*9.81 = .4 a
-T + .4*9.81 = .4 a
--------------------subtract
3 T =- .8*9.81 = 0
T = .8 * 9.81/3
go back and get a
v = a t
d = (1/2) a t^2