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Travka [436]
3 years ago
5

A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is

the normal force acting on the block?
Physics
1 answer:
blagie [28]3 years ago
4 0

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

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Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the
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Answer:

Part a)

a= 0.32 m/s^2

Part b)

F_c = 3.6 N

Part c)

F_c = 5.5 N

Explanation:

Part a)

As we know that the friction force on two boxes is given as

F_f = \mu m_a g + \mu m_b g

F_f = 0.02(10.6 + 7)9.81

F_f = 3.45 N

Now we know by Newton's II law

F_{net} = ma

so we have

F_p - F_f = (m_a + m_b) a

9.1 - 3.45 = (10.6 + 7) a

a = \frac{5.65}{17.6}

a= 0.32 m/s^2

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

F_c - F_f = m_b a

F_c = \mu m_b g + m_b a

F_c = 0.02(7)(9.8) + 7(0.32)

F_c = 3.6 N

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

F_p - F_f - F_c = m_b a

9.1 - 0.02(7)(9.8) - F_c = 7(0.32)

F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)

F_c = 5.5 N

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