Physics, 05.02.2020 08:00 angeladominguezgarci

Acave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. the lift is performed in three stages, each requiring a vertical distance of 10.0 m. (1) first, the initially stationary spelunker is accelerated to a speed of 4.40 m/s. (2) then he is then lifted at the constant speed of 4.40 m/s. (3) finally he is decelerated to zero speed. how much work is done on the 78.0 kg rescuee by the force lifting him during each stage? work done on rescuee during stage (1) kj work done on rescuee during stage (2) kj work done on rescuee during stage (3) kj

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Answer from: bestdiva2

(1) 8408 J = 8.4 kJ

We need to find the acceleration of the spelunker first, through the equation:

v^2-u^2=2ad

where

v = 4.40 m/s is the final speed

u = 0 is the initial speed

a is the acceleration

d = 10.0 m is the distance

Solving the equation for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{(4.40 m/s)^2-0}{2(10.0 m)}=0.97 m/s^2$

In order to find the magnitude of force F used to lift the spelunker, we have to apply Newton's second law:

$\sum F = ma\\F - mg = ma$

where (mg) is the weight of the spelunker, and a = 0.97 m/s^2. Solving for F, we find

F=m(g+a)=(78.0 kg)(9.81 m/s^2+0.97 m/s^2)=840.8 N

And so, the work done by the force during this stage is

W=Fd=(840.8 N)(10.0 m)=8408 J

(2) 7644 J = 7.6 kJ

The work done on the spelunker in this stage is

W=Fd

where F is the force applied on the spelunker to lift him, and d = 10.0 m is the vertical distance through which the spelunker is lifted.

In order to find the magnitude of F, we have to apply Newton's second law:

$\sum F = ma\\F - mg = ma$

where (mg) is the weight of the spelunker, and the acceleration is zero because he is moving at constant speed: so, a=0, and the equation becomes

$F-mg=0\\F=mg=(78.0 kg)(9.8 m/s^2)=764.4 N$

So, the work done is

W=(764.4 N)(10.0 m)=7644 J

3) 6895 J = 6.9 kJ

This stage is similar to stage (1); we find the deceleration using:

v^2-u^2=2ad

where

v = 0 m/s is the final speed

u = 4.40 is the initial speed

a is the acceleration

d = 10.0 m is the distance

Solving the equation for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{0-(4.40 m/s)^2}{2(10.0 m)}=-0.97 m/s^2$

In order to find the magnitude of force F used to lift the spelunker, we have to apply Newton's second law:

$\sum F = ma\\F - mg = ma$

where (mg) is the weight of the spelunker, and a = -0.97 m/s^2. Solving for F, we find

F=m(g+a)=(78.0 kg)(9.81 m/s^2-0.97 m/s^2)=689.5 N

And so, the work done by the force during this stage is

W=Fd=(689.5 N)(10.0 m)=6895 J

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

Answer from: Quest

root word, prefixes, and suffixes

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