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

Question

Nuetrik [128]

3 years 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] · Answer 1 · 2022-07-24T12:13:03+03:00

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

Arlecino [84] · Answer 2 · 2022-07-24T13:28:58+03:00

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

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<h3>Further explanation</h3>

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:

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