Respuesta :
We need first to use the formula F=m(a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be
m=2kg(both blocks)+500g(both ropes) → m=2.5kg
a=3.00m/s^2
g=9.8m/s^2
F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N
To calculate the tension at the top of rope 1 you need to use the formula T=m(a+g) so it will be T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N
We can now calculate the tension at the bottom of rope 1 using the formula: T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N
Now to find the tension at the top of rope 2 we do it like this:
T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2
m=2kg(both blocks)+500g(both ropes) → m=2.5kg
a=3.00m/s^2
g=9.8m/s^2
F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N
To calculate the tension at the top of rope 1 you need to use the formula T=m(a+g) so it will be T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N
We can now calculate the tension at the bottom of rope 1 using the formula: T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N
Now to find the tension at the top of rope 2 we do it like this:
T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2
The tension at the top end of rope 1 is 19.2 Newton
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Further explanation
Let's recall Impulse formula as follows:
[tex]\boxed {I = \Sigma F \times t}[/tex]
where:
I = impulse on the object ( kg m/s )
∑F = net force acting on object ( kg m /s² = Newton )
t = elapsed time ( s )
Let us now tackle the problem!
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Complete Question:
The figure shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s² by force F. The tension , in N , at the top end of Rope 1 is closest to ?
Given:
mass of block A = mass of block B = M = 1.0 kg
mass of rope 1 = mass of rope 2 = m = 250 g = 0.25 kg
acceleration of system = a = 3.0 m/s²
gravitational acceleration = g = 9.8 m/s²
Asked:
tension at the top end of rope 1 = T = ?
Solution:
We will use Newton's Law of Motion to solve this problem as follows:
The Whole System:
[tex]\Sigma F = ( 2m + 2M ) a[/tex]
[tex]F - W = ( 2m + 2M ) a[/tex]
[tex]F - ( 2m + 2M ) g = ( 2m + 2M ) a[/tex]
[tex]F = ( 2m + 2M ) ( a + g )[/tex]
[tex]F = ( 2(0.25) + 2(1.0) ) ( 3.0 + 9.8 )[/tex]
[tex]\boxed {F = 32 N}[/tex]
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Block A:
[tex]\Sigma F_A = Ma[/tex]
[tex]F - W_A - T = Ma[/tex]
[tex]F - Mg - T = Ma[/tex]
[tex]T = F - Mg - Ma[/tex]
[tex]T = F - M( g + a )[/tex]
[tex]T = 32 - 1.0 ( 9.8 + 3.0 )[/tex]
[tex]T = 32 - 12.8[/tex]
[tex]\boxed {T = 19.2 \texttt{ Newton}}[/tex]
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Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
- Average Speed of Plane : https://brainly.com/question/12826372
- Impulse : https://brainly.com/question/12855855
- Gravity : https://brainly.com/question/1724648
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Answer details
Grade: High School
Subject: Physics
Chapter: Dynamics