Question

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...

Answer 1

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.

Answer 2

yay i got it right thanks!