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madreJ [45]
3 years ago
14

The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

of the elevator stop the elevator shortly before it reaches the bottom of the shaft. If the elevator fell a distance of 100 m starting from rest, the heat that the brakes must dissipate to bring the elevator safely to rest is (A) 100 J (B) 1,000 J (C) 10,000 J (D) 100,000 J (E) 1,000,000 J
Physics
1 answer:
tester [92]3 years ago
4 0

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake  = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

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Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
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Answer:

1)4.7334J

2)225.4m/s

Explanation:

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Using the law of conservation of energy

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We can write eqn(1) as

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If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

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1) We can now calculate the total energy of the system after collision as

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= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

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Answer:

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