TEXTBOOK: GIANCOLI PHYSICS VOLUME 1 CHAPTER 5. PROBLEM 53 I have a question for

Question

TEXTBOOK: GIANCOLI PHYSICS VOLUME 1 CHAPTER 5. PROBLEM 53 I have a question for this problem for PartC. The question reads: What will a spring scale read for the weight of a 55kg womanin an elevator that moves (a) upwards with constant speed of 6.0ms,(b) downward with a constant speed of 6.0 m/s, (c) upward with acceleration of .33g, (d) downwardwith acceleration of .33g and (e) in free fall? For (a) and (b) i calculated FN= 5.4*10^2N However for C i became confused because i was looking atcramster's solution and it reads FN=m(1+.33)g FN= (55*1.33)*9.8m/s^2N =7.2*10^2N The only question i have is why 1 is added to theacceleration. If any clarification is needed please respond. If not. PLEASEHELP!! Thank you! I WILL RATE YOUR RESPONSES! TEXTBOOK: GIANCOLI PHYSICS VOLUME 1 CHAPTER 5. PROBLEM 53 I have a question for this problem for PartC. The question reads: What will a spring scale read for the weight of a 55kg womanin an elevator that moves (a) upwards with constant speed of 6.0ms,(b) downward with a constant speed of 6.0 m/s, (c) upward with acceleration of .33g, (d) downwardwith acceleration of .33g and (e) in free fall? For (a) and (b) i calculated FN= 5.4*10^2N However for C i became confused because i was looking atcramster's solution and it reads FN=m(1+.33)g FN= (55*1.33)*9.8m/s^2N =7.2*10^2N The only question i have is why 1 is added to theacceleration. If any clarification is needed please respond. If not. PLEASEHELP!! Thank you! I WILL RATE YOUR RESPONSES!

Explanation / Answer

We have writing Newton's second law for the vertical direction , withup as positive we have      F = FN-mg = ma            or FN = m(g+a) C) here a = +0.33g thus FN = m(g+0.33g)          =mg(1+0.33)           =mg*1.33            =1.33*55kg*9.8m/s2            = 7.2*102N e) here a = -0.33g so FN = m(g-0.33g)          =mg(1-0.33)          =0.67*55kg*9.8m/s2            =3.6*102N e) here a = -g thus FN = m(g+a)              = m(g-g)               = 0N            = 7.2*102N e) here a = -0.33g so FN = m(g-0.33g)          =mg(1-0.33)          =0.67*55kg*9.8m/s2            =3.6*102N e) here a = -g thus FN = m(g+a)              = m(g-g)               = 0N            =3.6*102N e) here a = -g thus FN = m(g+a)              = m(g-g)               = 0N

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