Applying Newton's Laws of Motion

[emma3001]

Homework Statement

Blocks X and Y of masses mx=5.12kg and my=3.22kg are connected by a fishing line passing over a frictionless pulley. Show that block X slides up the incline (35.7 degrees above the horizontal) with positive acceleration. Determine the magnitude of the acceleration. (0.273m/s2 is the answer)

Homework Equations

Ftension-Fg=Fnet=m total x a system
sin35.7=Fty/Ftens
Cos35.7=Ftx/Ftens
Fg=mg (Block Y)

The Attempt at a Solution

I want to find the Ftx= and FAy but i do not know the Ftens. I can calculate the gravitational force of Block Y, which is F=mg=31.6N. now i am trying to figure out how to calculate the net force which will give me acceleration, using the combined mass of the two blocks. What is confusing me is the incline and the fact that no tension force is given.

 

[emma3001]

please help me understand the situation at least! thx

 

[ogb p]

I would approach this problem by first identifying the relevant forces acting on the system. In this case, we have the force of gravity acting on both blocks, the normal force of the incline acting on block X, and the tension force in the fishing line connecting the two blocks.

Using Newton's Second Law, we can write the equation Fnet = ma, where Fnet is the net force acting on the system, m is the combined mass of the two blocks, and a is the acceleration of the system.

To determine the net force, we need to consider the forces acting on each block separately. For block X, we have the force of gravity pulling it down the incline and the normal force pushing it up the incline. The normal force can be calculated using the formula Fnormal = mgcosθ, where θ is the angle of the incline.

For block Y, we have the force of gravity pulling it down and the tension force pulling it up. Since the pulley is frictionless, the tension force will be equal to the weight of block X, which we can calculate using the formula Ftx = mgsinθ.

Now, we can write the equation for the net force as Fnet = (mX + mY)a, where mX and mY are the masses of blocks X and Y respectively.

Substituting in the forces we calculated for each block, we get:

Fnet = (mX + mY)a = (mXgcosθ - mYgsinθ) = (5.12kg)(9.8m/s²)(cos35.7°) - (3.22kg)(9.8m/s²)(sin35.7°) = 28.87N

Finally, we can solve for the acceleration by rearranging the equation to a = Fnet / (mX + mY) = 28.87N / (5.12kg + 3.22kg) = 0.273m/s².

Therefore, we can conclude that block X will slide up the incline with a positive acceleration of 0.273m/s².