#$&* course Phy 241 October 7 around 1:30pmI just took out general phy and prin phy problems so it would be easier for you to grade, instead of scrolling through all the problems together.
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Given Solution: `aThe weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx.. So the parachutist experiences a downward force of 540 N and an upward force of 620 N. Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is -540 + 620 N = 80 N. Your free body diagram should clearly show these two forces, one acting upward and the other downward. The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): PERFECT! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `quniv phy (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator. The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction. What is the net force on the fish when the balance reads 50 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** This wasn’t an assigned problem but I attempted it: Net force on fish at 50N: Fnet=ma You have 2 forces acting on the fish: the 50N upward force from the cable and the downward force of “mg” due to gravity. True weight of fish: 50N-mg=a M=50N/(a+g) =50N/(2.45m/s^2+9.807m/s^2) =50N/12.257m/s^2 =4.08g Balance will read 30N when: Fnet=ma 30N-mg=a 30N-(4.08g)(9.807m/s^2) = (-10N) A= (-10N)/4.08g =(-2.45m/s^2) so downward at 2.45m/s^2 When the cable holding the fish breaks, the balance will read: 0, simply because everything will be accelerating downward at the acceleration due to gravity, and therefore the balance won’t exert any force. #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Weight is force exerted by gravity. Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then F_net = m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did everything right, however, I had a few more sig figs, which made the answer differ slightly. I also said “4.08g” when it should’ve been 4.08kg."