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PolarNik [594]
2 years ago
9

Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no f

riction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?

Physics
2 answers:
GrogVix [38]2 years ago
5 0

Answer:

T<m2hh

Explanation:

Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no friction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?

Two masses  m1 and m2  is released and there is tension T in the string

a is the acceleration of the system

Therefore

For m1

T-m1g=m1a

T=m1(g+a)

m2

this is so because te weight is going downward wile tension in te rope is facing upward

m2g-T=m2a

T=m2(g-a)

from the equation ,

T<m2

tankabanditka [31]2 years ago
4 0

Answer:m_1g

Explanation:

Given

Two masses m_1  and m_2  is released and there is tension T in the string

Suppose a is the acceleration of the system

Therefore from Diagram

For m_1

T-m_1g=m_1a

T=m_1(g+a)------1

for m_2 body

m_2g-T=m_2a

T=m_2(g-a)-------2

From above two Equation it is said that Tension is greater than m_1g and less than m_2g

m_1g

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
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Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

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