a mass (m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m2), which hangs vertically.
If the acceleration is .098 m/s^2 and m1 is 1.0 kg how much must m1 must be to keep it at this acceleration
I got 99 kg
could you please tell me how to do this if i did it wrong... it\'s sort of really late at night and... well i\'m tired...
The total mass is (m1+m2).
We assume no friction between the horizontal surface, nor the pulley.
The force causing acceleration is due to gravity on the mass m2 only, thus the force is F=m2g.
Using Newton's second law, F=ma, we obtain
m2g = (m1+m2)0.098
Now solve for m1:
m1 = (m2g/0.098)-m2
= (m2*9.8/0.098)-m2
= 100m2-m2
= 99m2
Remark: the horizontal surface has to be very smooth for this to happen, i.e. accelerating very slowly.
yay i got it right thanks!
To find the value of m2 that would keep the system at the given acceleration, you can use Newton's second law of motion (F = ma). Additionally, you can utilize the relationship between the masses and the tension in the cord.
Let's denote the tension in the cord as T. Since the surface is smooth, there is no friction, so the tension force acting on m1 will be the only external force affecting its acceleration.
For mass m1:
F1 = m1 * a
Since tension T is acting in the opposite direction of the acceleration:
F1 = T
For mass m2, the gravitational force acts downward:
F2 = m2 * g
Using the relationship between the masses and tension:
T = F2 = m2 * g
Setting these two equations equal to each other:
m1 * a = m2 * g
Now, let's substitute in the known values:
1.0 kg * 0.098 m/s^2 = m2 * 9.8 m/s^2
Simplifying:
0.098 = 9.8 * m2
Dividing both sides by 9.8:
m2 = 0.098 / 9.8
Calculating:
m2 = 0.01 kg
Therefore, the mass of m2 would have to be approximately 0.01 kg to keep the system at an acceleration of 0.098 m/s^2.
It seems like there was an error in your calculation. Instead of 99 kg, the correct answer is 0.01 kg. Please double-check your calculations when you are well-rested.