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gayaneshka [121]

2 years ago

7

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

ind
a) The acceleration of the masses
1.8 m/s^2
2.4 m/s^2
3.2 m/s^2
4.1 m/s^2
b) The tension in the rope.
18 N
26 N
38 N
44 N

1 answer:

photoshop1234 [79]2 years ago

5 0

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

<em>T</em> - 25 N = (8 kg) (2.4 m/s²)

<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N

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