Coefficient of kinetic friction between the two blocks

AI Thread Summary
The coefficient of kinetic friction between the two blocks is 0.30, with a frictionless table and pulleys. The masses involved are m1=2 kg, m2=3 kg, and m3=10 kg. Initial calculations yield an acceleration of -6.15 m/s², which differs from the book's answer of -5.7 m/s². Adjusting the equations leads to a recalculated acceleration of approximately 5.76 m/s². The discrepancy highlights the importance of correctly applying the frictional force in the equations of motion.
Squeezebox
Messages
57
Reaction score
0

Homework Statement


The coefficient of kinetic friction between the two blocks is 0.30. The surface of the table and pulleys are frictionless.

m1=2 kg
m2=3 kg
m3=10 kg

Homework Equations


\SigmaF=ma
T2-m3g=m3a
T1-T2=m2a
friction-T1=m1a


The Attempt at a Solution


T2=m3g+m3a
T1=friction-m1a

friction-m1a-m3g-m3a=m2a

(friction-m3g)/(m1+m2+m3)=a
a=-6.15m/s2

The answer in my book say it is -5.7m/s2, though.
 

Attachments

  • blocks copy.jpg
    blocks copy.jpg
    6.5 KB · Views: 705
Physics news on Phys.org
Squeezebox said:

Homework Statement


The coefficient of kinetic friction between the two blocks is 0.30. The surface of the table and pulleys are frictionless.

m1=2 kg
m2=3 kg
m3=10 kg
The answer in my book say it is -5.7m/s2, though.
T2=m3g-m3a
T1=m1a+m1gμ
T2=T1+m1gμ+m2a
a=(m3g-m1gμ-m1gμ)/(m1+m2+m3)=5.7552
 
Thread 'Collision of a bullet on a rod-string system: query'

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...
View full post »
Thread 'A cylinder connected to a hanged mass'

Thread 'A cylinder connected to a hanged 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...
View full post »

Similar threads

Accelerating pulleys (3 pulleys and 3 masses)
Replies
5
Views
2K
Calculating tensions and acceleration
Replies
10
Views
2K
Block-pulley system and rotational dynamics
Replies
5
Views
5K
Find the acceleration of the system (2 blocks sliding on a table)
Replies
23
Views
2K
Three masses suspended on pulley
Replies
5
Views
7K
Newton's second law 3 pulley problem.
Replies
2
Views
2K
Newton's law problem: Pushing 2 stacked blocks on a horizontal table
Replies
13
Views
3K
Angular Acceleration and Linear Acceleration of a Pulley
Replies
4
Views
2K
Newton's Second Law (Pulley Sustem)
Replies
3
Views
2K
I have a question about a pulley and tension-related problem
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top