Question

What is the acceleration of mass `m`?[take g=10`m/s^2`]

Answer 1

The accerlation is approximately `0.31` m/`s^2`

Explanation:

For this problem I will assume that the pulley is massless and frictionless as it is not stated otherwise

Start by determining expressions for the net force of each object. When released, the object will clearly accelerate down the left incline, because the slope is greater. Thus

`F_"net right" = m_"total"a`

`F_t - mgsin(theta) - mgmucostheta = 2ma`

From the other one

`mgsin(alpha) - F_t - mgmucosalpha = 2ma`

Solving for tension in the first equation:

`F_t = 2ma + mgsintheta + mgmucostheta`

Substitute into the second .

`mgsinalpha - 2ma - mgsintheta - mgmucostheta - mgmucosalpha = 2ma`

Put in the values now

`mgsin(53˚) - mgsin(37˚) - mg(0.05)cos(53˚) - mg(0.05)cos(37) = 2ma`

`4a = gsin(53˚) - gsin(37˚) - g0.05cos(53˚) - g0.05cos(37˚)`

`a = 0.31 m/s^2`

Hopefully this helps!