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Shah 72
MHB
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I got the speed of p when q reaches the pulleys = 1m/s, a= 2m/s^2
Iam getting time = 0.8s for q(b)
I got the ans for this.skeeter said:You really need to figure out how to post a readable image ...
View attachment 11169
Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.
$Mg - T = Ma$
$T - mg = ma$
solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.
Iam still having doubts with q(b)skeeter said:You really need to figure out how to post a readable image ...
View attachment 11169
Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.
$Mg - T = Ma$
$T - mg = ma$
solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.
skeeter said:You really need to figure out how to post a readable image ...
View attachment 11169
Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.
$Mg - T = Ma$
$T - mg = ma$
solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.
The ans is 0.1s but the textbook says 0.9sskeeter said:when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.
$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$
Oh I got it. It will be a quadratic equation and I solve using quadratic formulaskeeter said:when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.
$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$
Thank you so much. t2= 0.4s so total time will be 0.9 sskeeter said:when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.
$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$
Connected particles in mechanics refer to two or more particles that are linked together by a physical connection, such as a string, rod, or spring. These particles may have different masses and can move in different directions, but their motion is affected by the connection between them.
The motion of connected particles is described using Newton's laws of motion and the principles of conservation of energy and momentum. These laws and principles help to determine the forces acting on the particles and how they will move in response to these forces.
A rigid connection in mechanics refers to a connection between particles that does not allow for any deformation or change in length. On the other hand, a non-rigid connection allows for some flexibility or change in length between the particles. This can affect the forces and motion of the connected particles.
The acceleration of connected particles can be calculated using Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. By considering the forces acting on each connected particle, the net force and resulting acceleration can be determined.
To analyze the motion of connected particles, one must first identify all the forces acting on the particles and determine the net force. Then, using the principles of conservation of energy and momentum, the equations of motion can be solved to determine the position, velocity, and acceleration of the particles at any given time. Graphs and diagrams can also be used to visualize the motion of connected particles.