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Anvisha [2.4K]
4 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:
Alborosie4 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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Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
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1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


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