Question

Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging mass is 40 kg. The coefficient of sliding friction is .30.

Answer 1

The system's tension is 616 N and acceleration is 5.6
`m / s^(2)`

Step-by-step explanation:

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

`F_(n e t)=m_(t o t) * a`

Where,

`F_(n e t)` is net force acting on body

`m_{\mathrm{tot}}` is mass of body

a is acceleration of body

Given values

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

`a=\frac{F_(n e t)}{m_{\mathrm{tot}}}=\frac{m * g}{m_{\mathrm{tot}}}`

Put the value for m = hanging mass = 40 kg and
`g=9.8 \mathrm{m} / \mathrm{s}^(2)`, we get

`a=(40 * 9.8)/(30+40)=(392)/(70)=5.6 \mathrm{m} / \mathrm{s}^(2)`

The tension in the ropes,
`T=(m * g)+(m * a)`

Here, m as hanging mass

T = tension, N or
`k g m / s^(2)`

m = mass, kg

g = gravitational force,
`9.8 \mathrm{m} / \mathrm{s}^(2)`

a = acceleration,
`m / s^(2)`

`T = (40 * 9.8)+(40 * 5.6) = 392+224 = 616 N`