Question

A cave 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 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?

Answer 1

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

$W_c = 6003.25 J$