Class Management Help HW 7-1 Begin Date: 3/13/2018 7:00:00 PM n Due Date: 3/29/2
ID: 2035443 • Letter: C
Question
Class Management Help HW 7-1 Begin Date: 3/13/2018 7:00:00 PM n Due Date: 3/29/2018 11:59-00 PM End Date: 5/1/2018 11:59:00 PM (25%) Problem 3: A block of mass m is initially at rest at the top of an inclined plane, which has a height of 4./ m and makes an angle of e 159 with respect to the horizontal. After being released, it is observed to be traveling at v-0.2 m/s a distance d after the end of the inclined plane as shown. The coefficient of kinetic friction between the block and the plane is ?,-O 1, and the coefficient of friction on the horizontal surface is ?,-02. initial Final and evant material. theexpertta.com "b l ?50% Part (a) What is the speed of the block. In meters per second. Just after it leaves the inclined plane ? ed ed Grade Summary bottom536 Late work % Late Potential 75% Submissions Attempts remaining 1 cos0 tan0 S 9HOME cotanO asino atanacotanOsih() coshOl tanhO cotanh0 (0% per attempt) END tailed view DegreesRadians Submit gave up Feedback: deduction per feedback Hints: foraOL deduction. Hints remasing Use the wok-energy thecrem, and be sure to inelude the dssipative es. What is the kinetie energy at inatial pount? What s the potential energy when the block reaches ground le eh What suga should the work due to friction have? Does at increase o decrease the final energy? 50% Part (b) Find the distance, d. In metersExplanation / Answer
The block's kinetic energy KE at the bottom of the plane = it's potential energy PE at the top of the plane minus the energy consumed by friction on the way down. That KE at the bottom of the incline is then consumed by friction moving along the surface.
At x=d: v² = 0.04
= Vi2 -2*a*d
where Vi is the velocity at the bottom of the plane, a is the negative acceleration (deceleration) due to friction and d is the distance.
0.2² = Vi² - 2*a*d
PE = m*9.8*4.1
= 40.18*m where m = mass
The length of the plane = 4.1/sin15 = 15.84m
The force of friction on the plane = mgcos15*0.1 = 0.9475*m N
The energy consumed sliding down the plane = 0.9475*m*15.84 = 15.0095*m J
Therefore the energy remaining at the bottom of the plane = (40.18-15.0095)*m
= 25.17047*m J
25.17047*m = 1/2 *m*V²
v=7.095 or v=7.1
(b) Find the distance d in meters.
Use Vf² = Vi² - 2*a*d
= 0.2² = 7.095² - 2*a*d
= a*d = 25.15
The force of friction on the surface = m*g*0.2 = 1.96*m
The deceleration = F/m = 1.96*m/m = 1.96m/s²
= d = 25.15/1.96
= 12.83 meters