Finding acceleration in two blocks, one hanging and other on board.

In summary, two blocks connected by a string have an upper block with an acceleration of 2.6 m/s^2 if the coefficient of kinetic friction between the block and the table is 0.10. The rope and pulley are most likely massless. The y-component of acceleration for the upper block is zero. The formula for acceleration is a=(m2g-f)/(m1+m2) and the final answer for acceleration is 7.84 m/s^2. The tension in the string is T=9.8+N and the normal force for the upper block is N=9.8.
  • #1
DrClone
4
0

Homework Statement


Two blocks are connected by a string as in the figure below.

ch05.P070.jpg


What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.10?

Express your answer to two significant figures and include the appropriate units.


Homework Equations



?

The Attempt at a Solution


Attempt: I drew a free-body diagram for two blocks.

m1 = mass one (2.0kg)
m2 = mass two (1.0kg)
∫= friction
T= tension
w= weight

Block 1: Fx=m1(a) = T -∫

Fy= m1(a)=n-m1g

Block 2: Fx=0

Fy= m2(a)=T-m2g

I would find the normal force for Block 1 Fy then plug it into friction. Then...I got lost.
 
Last edited:
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  • #2
are the rope and pulley massless?
 
  • #3
The rope and pulley are most likely massless.

The y-component of acceleration for block 1 should be zero. See how that works.
 
  • #4
This is what I got so far...am I correct?

f+m1a=m2g-m2a?

Then

m1a+m2a=m2g-f

THEN

a(m1+m2)=m2g-f

a=(m2g-f)/(m1+m2)

?
 
  • #5
I can't really follow your formula, but what did you get for acceleration once you solved?
 
  • #6
The first box:

x-forces:
T - K = ma
T - (0.1)(N) = 2a

y-forces:
N - w = o
N = w
N = 2g
N = 9.8

The second box:

T - w = m2a
T - m2g = 1a
T - 9.8 = a

Solve that.
 
  • #7
2.6 m/s^2

it looks like I answered my own question...LOL

Thanks lolcat, I forgot the negative in the second box.
 
  • #8
Not a problem, but I actually got 7.84 m/s2 lol. Anyone else want to check our answers?
 

Related to Finding acceleration in two blocks, one hanging and other on board.

1. How do you calculate the acceleration of two blocks, one hanging and the other on board?

The acceleration of two blocks can be calculated using Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the hanging block will experience a downward force due to gravity (its weight) and the block on board will experience an upward force from the string connecting it to the hanging block. By setting the net force equal to the mass of the system (the two blocks) multiplied by the acceleration, the acceleration can be solved for.

2. What is the difference between the acceleration of the hanging block and the block on board?

The acceleration of the hanging block will be equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. However, the acceleration of the block on board will be slightly less than this due to the tension in the string connecting the two blocks. This tension will provide an upward force on the block on board, causing it to experience a slightly smaller acceleration than the hanging block.

3. How does the mass of the blocks affect the acceleration?

According to Newton's Second Law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that as the mass of the blocks increases, the acceleration will decrease, and vice versa.

4. Can the acceleration of the blocks change during the experiment?

The acceleration of the blocks can change during the experiment if there is a change in the net force acting on the system. For example, if the hanging block is released, the net force on the system will decrease and the acceleration will change accordingly.

5. What factors can affect the accuracy of the calculated acceleration?

The accuracy of the calculated acceleration can be affected by factors such as friction between the blocks and the surface they are on, air resistance, and the precision of the measurements taken. It is important to control these variables as much as possible in order to obtain an accurate result.

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