Solving Mass in Horizontal Pulley System: 99 kg?

AI Thread Summary
To solve the problem of a mass on a horizontal surface connected to a hanging mass, the net force equation (net force = ma) is crucial. With an acceleration of 0.098 m/s² and m1 given as 1.0 kg, the tension in the rope must be considered constant. Free-body diagrams should be drawn for both masses to analyze the forces acting on them. The calculations suggest that m2 must be 99 kg to maintain the specified acceleration. Understanding the relationship between the two masses and their respective forces is essential for confirming the solution.
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Homework Statement



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

Homework Equations



net force = ma

The Attempt at a Solution



i got 99 kg
please help me... it's really late at night and I'm tired...
 
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Do what you always should do with mechanics problems: draw free-body diagrams, then write out Newton's second law for each body. Remember that the tension in a massless rope hung over a massless pulley is constant throughout the rope.
 
Draw free-body diagrams of each object and the forces acting on them. If two objects are connected by a single string, then they have the same tension.
 
Do you know if I got it right?
 
your asking for m1? It's already given. Am I right?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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