(590) Problem 5: A block of mass m is initially at rest at the top of an incline
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Question
(590) Problem 5: A block of mass m is initially at rest at the top of an inclined plane, which has a height of 6.8 m and makes an angle of e27 with respect to the horizontal. After being released, it is observed to be traveling atv 0.75 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 up-0.1, and the coefficient of friction on the horizontal surface isur0.2. initial Final Otheexpertta.com -? 50% Part (a) What is the speed of the block, in meters per second just after it leaves the inclined plane? 5000 Part (b) Find the distance, d' in meters Grade Summa Deductions Potential ry 096 100% 7 8 9 cos0 asin0 a tanO ( acosO Sin cotan0a atanac Submissions Attempts remaining:!S (290 per attempt) detailed view acotan0inh0 cotanhO END Degrees Radians N0 Submit Hint I give up!Explanation / Answer
Given,
h = 6.8 m ; theta = 27 deg ; v = 0.75 m/s ; up = 0.1 ; ur = 0.2
a)using trigonometry
L = h/sin(theta) = 6.8/sin27 = 14.98 m
the forces in x direction
Fnet = ma = mg sin(theta) - u m g cos(theta)
a = g (sin(theta) - u cos(theta))
a = 9.81 (sin27 - 0.1 x cos27) = 3.58 m/s^2
We know from eqn of motion
v^2 = u^2 + 2 a s
v = sqrt (2 x 3.58 x 14.98) = 10.4 m/s
Hence, v = 10.4 m/s
on the horizontal surface
a = ug = 0.2 x 9.81 = 1.96 m/s^2
v^2 = u^2 + 2 a s
s = (10.4^2 - 0.75^2)/2 x 1.96 = 27.45 m
Hence, d = 27.45 m
(if u take v = 0 ; d = 27.59 m )
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