A 78.7 kg astronaut working outside the space station is stationary and has the

Question

A 78.7 kg astronaut working outside the space station is stationary and has the tether line break. To get back to the ship the 5.3 kg wrench is tossed kg away from the space station at 4.7 m/s in order to recoil him back to the space station. The speed of the astronaut back to (hopefully) the space station is ______ m/s. If the net work done on a moving object is negative (as by friction), what happens to the object's kinetic energy?

When a ball is tossed straight up in the air and falls back into one's hand, does the force of gravity do negative or positive work on ball as it goes up?

A truck, whose weight is 2.4-times that of a car, is traveling at 3.7-times the speed of the car. What is the ratio of the kinetic energy of the truck to that of the car (KEtruck/KEcar)?

Explanation / Answer

By law of conservation of momentum

m1v2=m2v2

78.7*v1=5.3*4.7       => v1= 0.32m/s

When work is negative

-W= KEf-KEi

will be true only if KEf < KEi,

Thus we can say KE of the object decreases finally.

W=F*d

Wg= Fg*h*cos180 = -Fgh

Thus Wg is negative

Weight is proportional to the mass

mtruck =2.4*mcar

vtruck = 3.7*vcar

KEtruck = ½* mtruck* vtruck = ½*2.4*mcar*3.7*vcar = (2.4*3.7)*(1/2* mcar* vcar) = 8.88* KEcar

KEtruck / KEcar = 8.88

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