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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 12.0 m: (a) the initially stationary spelunker is accelerated to a speed of 2.90 m/s; (b) he is then lifted at the constant speed of 2.90 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 77.0 kg rescue by the force lifting him during each stage

Respuesta :

Answer:

a) 323.4J

b) 0J

c) -323.4J

Explanation:

a) W=Fd

F=ma

solve for acc. using kinematics

v^2=vo^2+2a(x)

8.41=2a(12)

4.205=a(12)

0.35=a

F=(77)(0.35)

F=26.95N

W=26.95*12...... W=323.4J

b) No acceleration, thus no force, thus no work!

c) W=Fd

F=ma

find acc. using kinematics: v^2=vo^2+2a(x)

0=(2.9^2)+2a(12)

0=8.41+2a(12)

-8.41=2a(12)

-4.205=a(12)

-0.35=a

F=(77)(-0.35)

F=-26.95N

W=(-26.95)(12)

W=-323.4J

Yes, work can be negative!

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