- only 5B. A 3.00 kg mass is initially traveling across a horizontal surface with a speed of 9.00 m/s. 1 then travels an

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Only 5b A 3 00 Kg Mass Is Initially Traveling Across A Horizontal Surface With A Speed Of 9 00 M S 1 Then Travels An 1 (39.7 KiB) Viewed 76 times

Only 5b A 3 00 Kg Mass Is Initially Traveling Across A Horizontal Surface With A Speed Of 9 00 M S 1 Then Travels An 2 (38.75 KiB) Viewed 76 times

- only 5B. A 3.00 kg mass is initially traveling across a horizontal surface with a speed of 9.00 m/s. 1 then travels another 1.50 m across a rough surface where the coefficient of kinetic friction is of the system is 0.400 before contacting a linear spring. a) With what speed does the mass initially contact the spring with? V? Mp = 0.400 Fn = my 395 29.4 1.50m fx 29.4 (0.400)-11.76 IF 11.96- me V- To = -kx ( 2x) X Koko m V ut of - QUm9 (150) + 6(1:50) -260-400) (21,4) (150) +* (4:50) Ax = 1.50 3 EF=4.70+ - 13. 42 m b) If upon reaching the spring, the mass travels 0.200 m as it is brought to rest, what must the spring constant be assuming that friction continues to act while the compression occurs. Note that you are using the answer from part 'a' above for the contact speed. If you were not confident with parta', you may use 5.00 ms for the contact speed. Imis Imus + 2 xx - 4K mg AX Thom to 200m 315) (123510.400)(2,4)(0.200m) V Smit -0.200 Stx-ts-fx ma Kolts х
SA. A 3.00 kg mass, initially at rest, is dropped from a height of 25.0 m above ground level a) What is the kinetic energy of the mass at the instant before contacting the ground? 2SM KE - a muf +-1%82.147*235,337 Okt = -mg By -- 6925)-7355 = As -39)) 418421-9)(-15) Vio Vp? 0--9 Ax=25 = 22. 14 mis vo VO b) The same 3.00 kg mass is now held against a compressed spring and released from rest. If the spring was initially compressed by 0.200 m from its equilibrium position and it leaves the spring with the same amount of kinetic energy as found in part 'a', what must the spring constant of the spring be? F = -K (0.200m) KE-955.27 29.40 = - D.200 0.00 985,275-( 25.000 25.009 25.009 K = - 147 FS- 29.40 V4 = 22. Hmis ag Ar = 22.142 - -25.009 2(98)