10 points for helping me with part one and two! THANKYOU!! Part 1 A 2.50 kg part
ID: 1681192 • Letter: 1
Question
10 points for helping me with part one and two! THANKYOU!! Part 1 A 2.50 kg particle is subject to a net force that varies withposition as shown in the figure. The particle starts moving atx = 0, very nearly from rest. x = 5.00m ________________m/s part 2- k so for this problem i already got a,b, and c, i just need HELP PLEASE WITH D A 5.60 kg block is setinto motion up an inclined plane with an initial speed ofv0 = 8.00 m/s. Theblock comes to rest after traveling 3.00 m along the plane, whichis inclined at an angle of 30.0° to the horizontal. (a) For this motion, determine the change inthe block's kinetic energy.J
(b) For this motion, determine the change in potential energy ofthe block-Earth system.
J
(c) Determine the frictional force exerted on the block (assumed tobe constant).
N
(d) What is the coefficient of kinetic friction?
(a) For this motion, determine the change inthe block's kinetic energy.
J
(b) For this motion, determine the change in potential energy ofthe block-Earth system.
J
(c) Determine the frictional force exerted on the block (assumed tobe constant).
N
(d) What is the coefficient of kinetic friction?
Explanation / Answer
mass m = 5.60 kginitial speed of v = 8.00m/s Final velocity v ' = 0 distance S = 3.00 m angle = 30.0°
Accleration of the block a = [ v ' ^ 2- v ^ 2 ] / 2S mass m = 5.60 kg
initial speed of v = 8.00m/s Final velocity v ' = 0 distance S = 3.00 m angle = 30.0° = -10.6666 m / s ^ 2 we know a = - [gsin - g cos] -10.666 =-4.9 + 8.487 from this coefficient of friction = 0.679
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