What is the acceleration of a block pulled by a force on a frictionless surface?

[BuBbLeS01]

A block of mass 8.0 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 33.0 N at an angle 31.1° above the horizontal. What is the magnitude of the acceleration of the block?

Ok I am not sure how to go about solving this problem. I drew a diagram and I added weight and normal force. But now I don't know how to start the problem?

 

[PhanthomJay]

What did you show for the applied force, try breaking it into x and y components.

 

[FedEx]

A block of mass 8.0 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 33.0 N at an angle 31.1° above the horizontal. What is the magnitude of the acceleration of the block?

Ok I am not sure how to go about solving this problem. I drew a diagram and I added weight and normal force. But now I don't know how to start the problem?

I am here again. See as there is no friction on the surface the problem becomes much easier.Resolve the force in x and y components.The x comp will cause the motion in the block.

 

[BuBbLeS01]

Thank you so much AGAIN! You guys are awesome...I am understanding it more now!

 

[BuBbLeS01]

Now I have to find the magnitude of the normal force. So I got the y component...so do I multiply that by mg?
y component= 33.0 sin 31.1

 

[FedEx]

Now I have to find the magnitude of the normal force. So I got the y component...so do I multiply that by mg?
y component= 33.0 sin 31.1

No there is no need for the y component as we don't need it. If there were friction then we would have to conisder it. Right now we have just one equation that is fcos31.1=ma

 

[PhanthomJay]

The problem now asks to calculate the normal force. Why would you multiply? Consider all the forces acting in the y direction and use Newton 1 in the y direction to calculate the normal force. What forces act on the block in the y direction?

 

[BuBbLeS01]

The problem now asks to calculate the normal force. Why would you multiply? Consider all the forces acting in the y direction and use Newton 1 in the y direction to calculate the normal force. What forces act on the block in the y direction?

Weight acts in the y direction. I used that fcos31.1=ma equation but it says its wrong?

 

[PhanthomJay]

Two parts to this problem if I understand correctly. The first asks you to calculate the acceleration of the block, which is in the x direction. Please show your work for that. You must brak up the applied force into its x componnet and then use F_net = ma.

The 2nd part acts you to calculate the normal force, which is the contact force that the floor exerts vertically up on the block, in the positive y direction. The weight of the block is another force that acts down on the block in the negative y direction, which you have identified. What is the other force that acts on the block in the y direction? Once you identify all three forces, then you apply Newton's law in the y direction to solve for the normal force. Hint: since the block is not moving in the y direction, what is its acceleraion in the y direction?

 

[Sabellic]

My guess is as follows:

If the floor is frictionless, then neither Friction Force nor Normal Force need be considered. As well, the block moves only one direction horizontally, and there is no vertical motion. (The block doesn`t move up and down when it is pulled). Together, these two facts mean that the only force acting on the block is the tension (pulling) on the rope.

Therefore, to calculate the acceleration of the block, one need only to take the x component (in other words, the horizontal component) of the force into consideration.

 

[FedEx]

My guess is as follows:

If the floor is frictionless, then neither Friction Force nor Normal Force need be considered. As well, the block moves only one direction horizontally, and there is no vertical motion. (The block doesn`t move up and down when it is pulled). Together, these two facts mean that the only force acting on the block is the tension (pulling) on the rope.

Therefore, to calculate the acceleration of the block, one need only to take the x component (in other words, the horizontal component) of the force into consideration.

Thats what i am trying to say.

 

[Sabellic]

Thats what i am trying to say.

Oooops. LOL. Yep, I just looked at the above posts and you had already said the exact same thing before.