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ANEK [815]
3 years ago
8

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

s performed in three stages, each requiring a vertical distance of 11.0 m: (a) the initially stationary spelunker is accelerated to a speed of 1.10 m/s; (b) he is then lifted at the constant speed of 1.10 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 78.0 kg rescue by the force lifting him during each stage?
Physics
1 answer:
bija089 [108]3 years ago
7 0

Answer

Mass m = 78 kg

Vertical height in each stage h = 11 m

(a).

Initial speed u = 0

Final speed v = 1.1 m / s

v^2=u^2 + 2 as

1.1^2 = 2 a \times 11

a = 0.055 m/s²

Work done

W_a= m g h + \dfrac{1}{2}mv^2

W_a= 78\times 9.8 \times 11 + \dfrac{1}{2} 78 \times 1.1^2

W_a = 8408.4 + 47.19

W_a = 8455.59 J

(b).Work done

W_b= mgh

W_b = 78× 9.8× 11

W_b= 8408.4 J

c)

Work done

W= m g h + \dfrac{1}{2}m(v_f-v_i)^2

Where V = final speed

               = 0

            v = 1.1 m / s

for deceleration a = -0.055 m/s²

now,

F_L = 56 (-0.055+9.8) = 545.72 N

W_c = 545.75 × 11

W_c = 6003.25 J

   

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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
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Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

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