Consider the system of masses at left. (2) Two masses are connected by a light string and hung over a light, frictionless pulley (meaning we can neglect the masses of the string and pulley, and any friction forces in the equations). We will specify that the positive x-direction is to the left over the pulley as shown. direction means that m, drops while m2 rises. (Now consider the opposite - what motion in the negative direction would mean.) +x Thus motion in the positive W, m Free body diagrams are also shown for both hanging masses. Using separate instances of Newton's 2nd Law for these two masses, derive a single theoretical equation that solves for the acceleration of the system in terms of the unknowns and g. Hint: the tension should not be one of these unknowns. m. W, (3) (2). Assuming we take several trials where the total mass m, +m2 is Consider the setup and the equation you derived in Problem istribute the masses between m, and m,. The equation you derived ---- ----►

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consider the setup from Problem(2). Assume ?1 =4.00kg. ?2 =10.00kg, and no friction.

(a)What is the magnitude of the acceleration of the system?

(b)Specify in what directionthe systemisaccelerating.

(c)What is the tension in the string?

Consider the system of masses at left.
(2)
Two masses are connected by a light string and
hung over a light, frictionless pulley (meaning we
++
can neglect the masses of the string and pulley, and
any friction forces in the equations). We will specify
that the positive x-direction is to the left over the
pulley as shown. Thus motion in the positive
direction means that m, drops while m2 rises. (Now
consider the opposite - what motion in the negative
direction would mean.)
m,
W,
afar
%3D
T
om.j
Free body diagrams are also shown for both hanging masses. Using
separate instances of Newton's 2nd Law for these two masses, derive
a single theoretical equation that solves for the acceleration of the
system in terms of the unknowns and g. Hint: the tension should not
be one of these unknowns.
+x:
m,
W,
Consider the setup and the equation you derived in Problem
(3)
(2). Assuming we take several trials where the total mass m, + m, is
constant, but we redistribute the masses between m, and m,. The equation you derived
as your answer to the Problem (2) would apply.
Transcribed Image Text:Consider the system of masses at left. (2) Two masses are connected by a light string and hung over a light, frictionless pulley (meaning we ++ can neglect the masses of the string and pulley, and any friction forces in the equations). We will specify that the positive x-direction is to the left over the pulley as shown. Thus motion in the positive direction means that m, drops while m2 rises. (Now consider the opposite - what motion in the negative direction would mean.) m, W, afar %3D T om.j Free body diagrams are also shown for both hanging masses. Using separate instances of Newton's 2nd Law for these two masses, derive a single theoretical equation that solves for the acceleration of the system in terms of the unknowns and g. Hint: the tension should not be one of these unknowns. +x: m, W, Consider the setup and the equation you derived in Problem (3) (2). Assuming we take several trials where the total mass m, + m, is constant, but we redistribute the masses between m, and m,. The equation you derived as your answer to the Problem (2) would apply.
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Step 1

Given:

The mass of the block m1 is 4 kg.

The mass of the block m2 is 10 kg.

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