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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 fâ— .

Verified Answer :

pkmnmushy

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

johanrusli

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:

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

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:

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

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

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

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

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

\(\boxed {F = 32 N}\)

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Block A:

\(\Sigma F_A = Ma\)

\(F - W_A - T = Ma\)

\(F - Mg - T = Ma\)

\(T = F - Mg - Ma\)

\(T = F - M( g + a )\)

\(T = 32 - 1.0 ( 9.8 + 3.0 )\)

\(T = 32 - 12.8\)

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

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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