Calculating Speed of a Moving Block with Friction and Pulley System

[dragonladies1]

1. Homework Statement
A block of mass m1 = 280 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic friction between the block and the plane is μk = 0.10. The block is attached to a second block of mass m2 = 220 g that hangs freely by a string that passes over a frictionless and massless pulley. Find its speed when the second block has fallen 30.0 cm.
cm/s



2. The attempt at a solution

Mass 1 Mass 2
Fground=280*9.81=2746.8 N Fm2 ground=220*9.81=2158.2 N
Fgy=Fgcos(30)
=433.714 N
Fgx=Fgsin(30)
=250.405 N
Fkinetics=(.1)*(250.405)
=25.04 N
Fx= 250.405-25.04= 225.36N

Fx=max
225.36=280*ax
ax=.805 m/s2

It would be greatly appreciated if someone could please help!

Thank you!

 

[songoku]

Fkinetics=(.1)*(250.405)
=25.04 N

friction = normal force x coefficient of friction, and the normal force is not equal to the force on x-direction.

Fx=max
225.36=280*ax
ax=.805 m/s2

There are total 3 forces on the x-direction. You are one less

 

[dragonladies1]

So in the x direction I have...

For the second mass the F_g=2158.2N

and on the first mass F_gx=250.405 N
and friction which= 433.714*.1=43.37 N

Would it be:

2158.2+250.405-43.7=ma_x

2364.905=280a_x

or is "m" both of the masses and acceleration is squared?

 

[songoku]

Hi dragonladies

You have to convert the mass to kg first.

So in the x direction I have...

For the second mass the F_g=2158.2N

The second mass doesn't have component on x-direction, only y-direction

and on the first mass F_gx=250.405 N

and friction which= 433.714*.1=43.37 N

Don't know how you got his value..

After you find the force components and friction, set the equation using Newton's second law then find acceleration. Finally,use kinematics to find the speed.