A 5.80?kg block is set into motion up an inclined plane with an initial speed of
ID: 1903452 • Letter: A
Question
A 5.80?kg block is set into motion up an inclined plane with an initial speed of vi = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of ? = 30.0? to the horizontal.
(a) For this motion, determine the change in the block's kinetic energy. ______________J
(b) For this motion, determine the change in potential energy of the block?Earth system. ________________J
(c) Determine the friction force exerted on the block (assumed to be constant)._________________N enter a number ???
(d) What is the coefficient of kinetic friction? _______________? enter a number..?
A 5.80?kg block is set into motion up an inclined plane with an initial speed of vi = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of ? = 30.0? to the horizontal. (a) For this motion, determine the change in the block's kinetic energy. ______________J (b) For this motion, determine the change in potential energy of the block?Earth system. ________________J (c) Determine the friction force exerted on the block (assumed to be constant)._________________N enter a number ??? (d) What is the coefficient of kinetic friction? _______________? enter a number..?Explanation / Answer
A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s.? The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine (a) the change in the blocks kinetic energy, (b) the change in the potential energy of the block-Earth system, and (c) the friction force exerted on the block (assumed to be constant). a, delta KE = .5*M*v^2= 160 N*M b, PE = MGH = 5*8.8066*3 sin 30 = 73.5495 N*M c, Friction Force: DE = KE-PE= 86.4505 = F * x = 3 F Force= 28.8 N Friction force = Force - Fg = 28.8 - MGsin30 = 4.3 Coefficient of friction = 4.3/(5*g*cos(30)) = 0.10
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