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Nuetrik [128]
1 year ago
12

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 2

50
g. the entire assembly is accelerated upward at 3.0 m/s2 by force fâ— .

Physics
2 answers:
lutik1710 [3]1 year ago
7 0
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 
<span>m=2kg(both blocks)+500g(both ropes) → m=2.5kg </span>
<span>a=3.00m/s^2 </span>
<span>g=9.8m/s^2 </span>
<span>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 </span>T=m(a+g) so it will be <span>T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N 
</span>We can now calculate the tension at the bottom of rope 1 using the formula: <span>T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N 
</span>Now to find the tension at the top of rope 2 we do it like this: 
<span>T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2</span>
Arlecino [84]1 year ago
7 0

The tension at the top end of rope 1 is 19.2 Newton

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Complete Question:</u>

<em>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 ?</em>

<u>Given:</u>

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

<u>Asked:</u>

tension at the top end of rope 1 = T = ?

<u>Solution:</u>

<em>We will use Newton's Law of Motion to solve this problem as follows:</em>

<u>The Whole System:</u>

\Sigma F = ( 2m + 2M ) a

F - W = ( 2m + 2M ) a

F - ( 2m + 2M ) g = ( 2m + 2M ) a

F = ( 2m + 2M ) ( a + g )

F = ( 2(0.25) + 2(1.0) ) ( 3.0 + 9.8 )

\boxed {F = 32 N}

\texttt{ }

<u>Block A:</u>

\Sigma F_A = Ma

F - W_A - T = Ma

F - Mg - T = Ma

T = F - Mg - Ma

T = F - M( g + a )

T = 32 - 1.0 ( 9.8 + 3.0 )

T = 32 - 12.8

\boxed {T = 19.2 \texttt{ Newton}}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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