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Anvisha [2.4K]
3 years ago
6

Suppose you find yourself in your friend's third floor apartment building when you smell smoke coming from outside the door. you

open the door an see nothing but flames protruding from the hallway. Realizing that your only means of escape is through the window of the apartment, your freind agrees to donate many pairs of nylons which you tie together to form a makeshift rope. this nylon rope(no pun intended) can withstand a maximum tension of 500 N without breaking. you have a mass of 70Kg on this particular day.
a) why can't you slide down the rope at a near constant speeed?
b) What is the smallest prssible acceleration you must achieve in order to slide down the rope without it breaking? ans:2.9m/s square
Physics
1 answer:
Alborosie3 years ago
8 0

(a) Because the tension in the rope would be larger than the maximum value allowed

When you are sliding down the slope, there are two forces acting on you:

- Your weight, W=mg, acting downward, where

m = 70 kg is your mass

g = 9.8 m/s^2 is the acceleration of gravity

- The tension in the rope, T, acting upward

Therefore, the equation of motion is

T-mg=ma

where a is the acceleration.

If you want to slide at constant speed, then the acceleration must be zero:

a = 0

And so the equation becomes

T-mg=0

This means that the tension in the rope must be:

T=mg=(70 kg)(9.8 m/s^2)=686 N

Which is larger than the maximum tension allowed in the rope (500 N), so the rope will break.

(b) 2.66 m/s^2 downward

Again, we must refer to the equation of the forces:

T-mg=ma

In this case, we want the tension in the rope to be the maximum allowed value,

T = 500 N

the other data are

m = 70 kg is your mass

g = 9.8 m/s^2 is the acceleration of gravity

Substituting into the equation, we can find the corresponding value of acceleration:

a=\frac{T-mg}{m}=\frac{500-(70)(9.8)}{70}=-2.66 m/s^2

where the negative sign means the acceleration is downward.

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Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that th
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A running student has half the kinetic energy that his brother has. The student speeds up by 1 m/s, at which point he has the sa
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Answer:

V = (√2) + 1) m/s

Explanation:

Let the mass and speed of the running student be M and V respectively.

We are told that when he speeds up by 1 m/s, he has the same kinetic energy as his brother.

Thus, his speed at which he mow has the same kinetic energy as his brother is (V + 1) m/s

Now, we are told that the mass of the student is twice as large as that of his brother. Thus, his brother's mass is; M/2

Since kinetic energy is given by the formula K.E = ½mv²

Therefore, since we want to find the original speed of both students and that the initial condition says that the running student had half the kinetic energy of the brother, we now initial condition as;

½MV²= ½(½(M/2)V²) - - - - (eq 1)

Since he has sped up by 1 m/s, and has a kinetic energy now equal to that of his brother, we have;

(½M(V + 1)²) = (½(M/2)V²) - - - - (Eq2)

Dividing eq 1 by eq 2 gives;

V²/(V + 1)²= 1/2

Taking square root of both sides gives;

V/(V + 1) = 1/√2

Cross multiply to give;

(√2)V = V + 1

(√2)V - V = 1

V((√2) - 1) = 1

V = 1/((√2) - 1)

Simplifying this using surfs gives;

V = [1/((√2) - 1)] × ((√2) + 1))/((√2) + 1))

V = ((√2) + 1))/1

V = (√2) + 1) m/s

8 0
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