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:

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

$m_2=m_1\times \frac{g+a}{g-a}$

$6=\frac{g+a}{g-a}$

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

$50=7a$

$a=\frac{50}{7}$

$a=7.142\ m/s^2$

Answer 2

Answer:

Answer:

7 m/s^2

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.

Explanation: