What is the mass of the other block in this block and pulley system?

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SUMMARY

The mass of the unknown block in the block and pulley system can be determined using Newton's second law and kinematic equations. Given a 100 kg block that takes 5.60 seconds to fall 1 meter, the acceleration can be calculated using the equation \(X_f = X_i + V_i(T_f - T_i) + 0.5(a)(T_f - T_i)^2\). The tension in the string remains constant, allowing the application of \(100g - T = 100a\) for the known mass and a similar equation for the unknown mass. Solving these equations yields the mass of the other block.

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Homework Statement


The 100 kg block in figure takes 5.60 s to reach the floor after being released from rest.
What is the mass of the other block. The distance between the 100 kg block and the floor is 1m.


Homework Equations


f=ma
Xf=Xi+Vi(Tf-Ti)+(.5)(a)(Tf-Ti)^2


The Attempt at a Solution


?
 
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Both blocks must move with the same acceleration. Tension in the string is the same. Apply the Newton's second law for both the masses.
Like for the known mass
100*g - T = 100*a.
Similarly write down the equation for the unknown mass. Solve for T.
Distance moved by the block is given and time is given. Using the relevant equation find the acceleration. Using T and a, find the unknown mass.
 
rl.bhat said:
Both blocks must move with the same acceleration. Tension in the string is the same. Apply the Newton's second law for both the masses.
Like for the known mass
100*g - T = 100*a.
Similarly write down the equation for the unknown mass. Solve for T.
Distance moved by the block is given and time is given. Using the relevant equation find the acceleration. Using T and a, find the unknown mass.

k thanks that did the trick
 

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