A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 995 and 1450 N, respectively. The acceleration of the cable is 0.620 m/s2, upward.
(a) What is the tension in the cable below the worker?
N
(b) What is the tension in the cable above the worker?
N
The space probe Deep Space I was launched on October 24, 1998. Its mass was 474 kg. The goal of the mission was to test a new kind of engine called an ion propulsion drive. This engine generated only a weak thrust, but it could do so over long periods of time with the consumption of only small amounts of fuel. The mission was spetacularly successful. At a thrust of 53 mN how many days were required for the probe to attain a velocity of 895 m/s (2002 mi/h), assuming that the probe started from rest and that the mass remained nearly constant?
One Response
#1 – No picture but I assume the worker is above the crate. The problem states that this whole mess is accellerating upward, so there must be a net force in that direction. If you imaging “cutting” the rope above the crate and doing a free body diagram, the sum of your forces would be “T” up (which you don’t know), and the weight of the crate down (1450 N). If you take up as the “+” direction, the net force causing the accelleration of the crate is Fnet = T-1450. Then use the F=ma to solve for “T” where “F” is T-1450, m is the weight of the crate (i.e. the mass you are accellerating) and “a” is given. Solve for T. I get T=2349 N. Same approach for the tension above the worker, only now your Fnet includes the weight of both the crate and the worker. I get T=3961N. BTW this unsafe lifting scenario would never occur in the real world, at least not in the US…
#2 – The spaceship problem is quite simple: just use V = A * T, where V=895 m/s & A=53 mN=0.053N. Solve for T but watch your units. I get a little over 8 million seconds, or 92.6 days.