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

  • Thread starter Thread starter candykane91
  • Start date Start date
  • Tags Tags
    Block Pulley
AI Thread Summary
The problem involves a 100 kg block that takes 5.60 seconds to fall 1 meter, prompting the need to determine the mass of the other block in the pulley system. The key equations used are Newton's second law and kinematic equations to find acceleration. Both blocks experience the same acceleration, and the tension in the string remains constant. By calculating the acceleration with the given distance and time, the unknown mass can be derived using the tension and acceleration equations. The discussion concludes with a confirmation that the approach successfully solved the problem.
candykane91
Messages
7
Reaction score
0

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


?
 
Physics news on Phys.org
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
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top