Question

A 1.0-kg mass (mA) and a 6.0-kg mass (mB) are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. Find the acceleration of the larger mass.

Answer 1

Answer:

7 m/s^2

Step-by-step explanation:

mA = 1 kg

mB = 6 kg

Let a be the acceleration and T be the tension in the string

By use of Newton's second law

T - mA g = mA x a .... (1)

mBg - T = mB x a ..... (2)

Adding both the equations

(mB - mA) g = (mA + mB) x a

(6 - 1) x 9.8 = (6 + 1) x a

7 a = 9.8 x 5

a = 7 m/s^2

Thus, the acceleration in the system is 7 m/s^2.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given

`m_1=1\ kg`

`m_2=6\ kg`

Pulley mass String

For
`m_1=1\ kg`

`T-m_1g=m_1a`

`T=m_1(g+a)`

for other body
`m_2`

`m_2g-T=m_2a`

`T=m_2(g-a)`

Equating value of Tension

`m_2=m_1* (g+a)/(g-a)`

`6=(g+a)/(g-a)`

`6(10-a)=10+a`

`50=7a`

`a=(50)/(7)`

`a=7.142\ m/s^2`