What will be the horizontal acceleration

AI Thread Summary
A 4.00-kg block is sliding up a 30-degree incline with a coefficient of kinetic friction of 0.700, and a 50-N horizontal force is applied. The initial calculations yielded an acceleration of 6.65 m/s², while the correct answer is 4.40 m/s². The error arose from treating the 50 N force as acting parallel to the incline instead of horizontally. To solve correctly, the horizontal force must be decomposed into components parallel and perpendicular to the incline, similar to how the weight is treated. Properly applying these principles will lead to the accurate calculation of the block's acceleration.
JaZZyCooL
Messages
7
Reaction score
0

Homework Statement


a 4.00-kg block is sliding up the plane on a 30 degree incline. If the coefficient of kinetic friction between the block and the inline is .700. What will be the acceleration of the block if a 50-N horizontal force pushes on the block?

Homework Equations



50 - mgsin(30) - Uk*mg*cos(30) = ma

The Attempt at a Solution


[/B]
I am getting answer 6.65 something whereas the correct answer is 4.40 m/s.
I suck at this topic. This is the weakest point I would appreciate your help.
 
Last edited by a moderator:
Physics news on Phys.org
JaZZyCooL said:

Homework Statement


a 4.00-kg block is sliding up the plane on a 30 degree incline. If the coefficient of kinetic friction between the block and the inline is .700. What will be the acceleration of the block if a 50-N horizontal force pushes on the block?

Homework Equations



50 - mgsin(30) - Uk*mg*cos(30) = ma

The Attempt at a Solution


[/B]
I am getting answer 6.65 something whereas the correct answer is 4.40 m/s.
I suck at this topic. This is the weakest point I would appreciate your help.

It seems like your setup is close. can you post all of your work so we can check it?
 
berkeman said:
It seems like your setup is close. can you post all of your work so we can check it?

50 - 4.0*9.8*sin(30)-4.0*0.7*9.8*cos(30) = 50 - 23.76 = 26.64
26.64/4.00 = 6.56 m/s^2
 
JaZZyCooL said:
50 - mgsin(30) - Uk*mg*cos(30) = ma
Looks like you've worked the problem for the case where the 50 N force acts parallel to the incline. But the force is acting horizontally.
 
So what will be the equation if the force is acting horizontal?
 
When you dealt with the weight mg, you broke it into components parallel and perpendicular to the slope. Do the same for the 50 N force.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top