# ch7 - 25(a The net upward force is given by F FN(m M g =(m...

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2 5 . (a) The net upward force is given by () N FF mMg mM a +−+ =+ where m = 0.250 kg is the mass of the cheese, M = 900 kg is the mass of the elevator cab, F is the force from the cable, and 3.00 N N F = is the normal force on the cheese. On the cheese alone, we have 2 3.00 (0.250)(9.80) 2.20 m/s 0.250 N Fm g m a a −= ¡ == . Thus the force from the cable is 4 ( ) 1 . 08 10 N N M a gF +−= × , and the work done by the cable on the cab is 44 1 (1.80 10 )(2.40) 2.59 10 J. WF d × = × (b) If 92.61 kJ W = and 2 10.5 m d = , the magnitude of the normal force is 4 2 9.261 10 (0.250 900)(9.80) 2.45 N. 10.5 N W M g d × + =

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3 3 . (a) This is a situation where Eq. 7-28 applies, so we have Fx = 1 2 kx 2 ¡ (3.0 N) x = 1 2 (50 N/m) x 2 which (other than the trivial root) gives x = (3.0/25) m = 0.12 m. (b) The work done by the applied force is W a = Fx = (3.0 N)(0.12 m) = 0.36 J. (c) Eq. 7-28 immediately gives W s = – W a = –0.36 J. (d) With K f = K considered variable and K i = 0, Eq. 7-27 gives K = Fx 1 2 kx 2 . We take the derivative of K with respect to x and set the resulting expression equal to zero, in order to find the position x c which corresponds to a maximum value of
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ch7 - 25(a The net upward force is given by F FN(m M g =(m...

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