A light rope passing over a frictionless pulley connects two objects. One object has a mass of 10kg. The tension in the rope is 90N. Calculate the acceleration and the mass of the second object.

Question

Object 1
Fg-Ft = ma
100 - 90 - 10*a
10/10 = a
a = 1m/s^2

Object 2
Ft-Fg = ma
90-m10 = ma

So I substituted 1 for acceleration

90-m10 = 1m

I still don't know what to do now.

Another question -
A 30 N force is pushing the 10kg object down the slope. The coefficient of friction between the object and the surface is 0.625. The slope is 37 degrees to the horizontal.

Calculate the acceleration of the object.

Perpendicular - balanced forces
EF = 0
Fn - Fgperp = 0
Fn - 100cos37 = 0
Fn = 79.86N

Parallel motion - unbalanced forces
EF = ma
Fa - Ffr + Fg// = ma
30 - (0.625)*(79.86) + 100Sin37 = 10a
a = 4.03m/s^2

Is that right?

I don't know what to do next. Help?

Answer 1

You have it.
Solve that last equation for the mass.
I would strongly recommend that you keep the units in your equations--especially ones that are challenging.

90N - m *(10 m/s^2) = m * (1 m/s^2)
90N = m (11 m/s^2)
(90N)/(11 m/s^2) = 8.18kg

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Your answer to the second question look OK.