two blocks of masses m1=4.5 kg and m2=6.5 kg resting on a frictionless surface are connected by a light inextensible cord. A horizontal force F of 33.0 N directed to the right is applied to the block with m1 and a horizontal force F of 11N directed to the left is applied to the block with m2. find the acceleration of the system and the tension in the rope.
Well, well, well! Looks like we've got ourselves a physics problem here. Don't worry, I'm here to clown around and make it fun!
To find the acceleration of the system, we need to apply some good old Newton's second law. The net force acting on the system is equal to the total mass times acceleration (F=ma).
For m1, the force applied to the right is 33.0 N, and for m2, the force applied to the left is 11 N. Let's combine those forces: 33 N (to the right) - 11 N (to the left) = 22 N. This 22 N is the net force acting on the system.
Now, we need to find the total mass of the system. It's just m1 + m2, so 4.5 kg + 6.5 kg = 11 kg.
Finally, we can put it all together: net force = total mass x acceleration. Plugging in the numbers, 22 N = 11 kg x acceleration. Solving for acceleration, we get:
acceleration = 22 N / 11 kg = 2 m/s².
So, the acceleration of the system is 2 m/s². Now, let's move on to the tension in the rope.
Since the blocks are connected by an inextensible cord, they will have the same acceleration. So, we can focus on just one block (let's say m1).
Applying Newton's second law to m1, we have the force applied to the right (33 N) minus the tension in the rope equals the mass of m1 times acceleration. That can be written as:
33 N - T = 4.5 kg x 2 m/s²,
where T is the tension in the rope. Now, let's solve for T:
33 N - T = 9 kg m/s²,
T = 33 N - 9 kg m/s²,
T = 24 N.
Therefore, the tension in the rope is a whopping 24 Newtons!
That's it! We've successfully solved the problem while clowning around a little. Hope you found it amusing, and remember, physics can be fun too!
To find the acceleration of the system and the tension in the rope, we can apply Newton's second law of motion to each of the blocks and then solve the equations simultaneously.
1. Start by writing the equation of motion for the block with mass m1:
ΣF₁ = m₁ * a (Equation 1)
Here, ΣF₁ represents the net force acting on m₁, and a is the acceleration of the system.
2. The net force acting on m₁ is the applied force F₁ − the tension in the rope:
ΣF₁ = F₁ − T (Equation 2)
3. Similarly, write the equation of motion for the block with mass m2:
ΣF₂ = m₂ * a (Equation 3)
4. The net force acting on m₂ is the applied force F₂ − the tension in the rope:
ΣF₂ = T − F₂ (Equation 4)
5. Set up a system of equations by substituting equations 2 and 4 into equations 1 and 3, respectively:
F₁ − T = m₁ * a (Equation 5)
T − F₂ = m₂ * a (Equation 6)
6. Substitute the given values into the equations:
F₁ = 33.0 N
F₂ = 11.0 N
m₁ = 4.5 kg
m₂ = 6.5 kg
7. Now, solve equations 5 and 6 simultaneously to find the values of a and T.
From equation 5:
F₁ − T = m₁ * a
33.0 − T = 4.5 * a (Equation 7)
From equation 6:
T − F₂ = m₂ * a
T − 11.0 = 6.5 * a (Equation 8)
8. Rearrange equation 7 to solve for T:
T = 33.0 − 4.5 * a (Equation 9)
9. Substitute equation 9 into equation 8 and solve for a:
33.0 − 4.5 * a − 11.0 = 6.5 * a
22.0 = 11.0 * a
a = 22.0 / 11.0
a = 2.0 m/s²
10. Substitute the value of a back into equation 9 to find T:
T = 33.0 − 4.5 * a
T = 33.0 − 4.5 * 2.0
T = 33.0 − 9.0
T = 24.0 N
Therefore, the acceleration of the system is 2.0 m/s², and the tension in the rope is 24.0 N.
So this gives you the mass and force of m1, use Newton's second law (F=m*a) to get the acceleration.... 33.0/4.5kg=7.33 m/s^2
I drew this situation as
[m1]----[m2] since the applied force was to the right the --- tension force becomes limp and should =0
[m1]>~~_[m2] it's easier to imagine pulling a rope, if you take a step inward the rope becomes limp and has no tension force present anymore.
Hope this helps!!
the mass of the system is 11 kg ... 6.5 + 4.5
the net external force is 22 N to the right ... 33 - 11
so the acceleration is ... f / m = 22 / 11 = 2 m/s^2 to the right
the tension in the rope combines with the 11 N force
... to give m2 the 2 m/s^2 acceleration
... t - 11 = 6.5 * 2