Question

. A 5 kg turkey and a 3.5 kg turkey are suspended on opposite sides of a single pulley.

Answer 1

Answer:

a) 40.3725 N

b) $a=1.729 m/s^2$

Explanation:

mass of turkey 1 M= 5 kg

mass of turkey 2 m= 3.5 kg

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

then from FBD in the attachment

T-mg=ma.................1

Mg-T=Ma..................2

from 1 and 2 we find a as

$a= \frac{Mg-mg}{M+m}$

now substituting the values we get

$a= \frac{5g-3.5g}{5+3.5}$

$a= \frac{1.5g}{8.5}$

$a=1.729 m/s^2$

now from equation 1 we can calculate Tension T as

T-mg=ma

T= m(a+g)

T= 3.5(1.729+9.81)= 40.3725 N