Question

# Two masses of 3.00 kg and 5.00 kg are connected by a light stringthat passes over a frictionless pulley, and was shown in Figure5.15a. Figure 5.15a (a

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Two masses of 3.00 kg and 5.00 kg are connected by a light stringthat passes over a frictionless pulley, and was shown in Figure5.15a.
Figure 5.15a
(a) Determine the tension in the string. N
(b) Determine the acceleration of each mass.
m/s2 upwards (3.00 kg mass)
m/s2 downwards (5.00 kg mass)
(c) Determine the distance each mass will move in the firstsecond of motion if they start from rest.
m (3.00 kg mass)
m (5.00 kg mass)
i don't understand how to do like any of this problem, someone please help me!!!!!

2021-01-11
Total forces acting on both masses together mustbe: forward, Mg backward,mg
The masses move together, so they have the same acceleration. So total force = total $$\displaystyle{m}{a}{s}{s}\times{a}$$
Mg - mg = ( M + m ) a
$$\displaystyle{a}={\frac{{{g{{\left({M}-{m}\right)}}}}}{{{\left({M}+{m}\right)}}}}={\frac{{{9.8}\times{2.00}}}{{{8.00}}}}={2.45}\frac{{m}}{{s}}$$ this is the acc for both Tension? consider just the smaller mass. total force = ma
T - mg = ma
T = mg + ma = m ( g + a) = $$\displaystyle{3.00}\times{\left({9.8}+{2.45}\right)}$$ = 36.75 Newtons
And the distance? They move the same distance. This can befound by using dist = $$\displaystyle{\frac{{{1}}}{{{2}}}}{a}{t}^{{{2}}}={\frac{{{1}}}{{{2}}}}\times{9.8}\times{1}^{{{2}}}={4.9}{m}{e}{t}{e}{r}{s}$$ they bothmove this distance in the first second