A surface is constructed as shown below with a horizontal surface connected to a
ID: 1344336 • Letter: A
Question
A surface is constructed as shown below with a horizontal surface connected to an incline with a downward slope of 30 degrees between them. The coefficient of static friction on the surface is 0.24. The coefficient of kinetic friction on the surface is 0.15.
The first box with a mass of 5.0 kg sits at rest on the horizontal surface 10. cm to the left of the start of the downward slope (as shown below). A second box with a weight of 19.6 N and a momentum of 3.0 kg*m/s moving left to right hits the first box elastically.
1. Assuming that the interaction of the collision lasts 0.1 s, what is the force that each box feels as a result of the collision? Remember that force has both magnitude AND direction.
2. What is the final position of both of the boxes? You may assume that overcoming static friction on the 5.0 kg box requires no work because it occurs over such a small distance. Be sure to define the coordinate system for the positions you provide and prove that the 5.0 kg box moves at all.
3. Draw a Free-Body Diagram
Explanation / Answer
the moment of second block moving with a 3 kg-m/s
weight=19.6 N
mass of m2 =2 kg
m1 = 5 kg is at rest.
F = force
t= time
d=distance
1)
from the conservation of momentum
3 = m1V1 + m2V2
3 = 5V1+2V2
from the conservation of energy
(1/2)(2)(3/2)2 = 0.5m1V12 + 0.5m2V22
2.25 = 2.5V12 + V22
we know,
V2 = 1.5 -2.5 V1
solving we get
V1 = 0.8571 m/s
V2 = -0.64275 m/s (negative sign means the direction is opposite)
Force on the block 1
We know that
impulse = change in momentum
F*t =( 5*0.8571 - 0 )
F = 4.2855/0.1
= 42.855 N
force on mass 2
F*t = change in momentum
F = (-.64572*2 - 3)/0.1
= -42.855 N
2) For final position of masses
Mass m1 = 5 kg
The K.E after collision = (1/2)m1V12
= 1.836 J
Friction force = umg
= 0.15*9.81
friction f*d = K.E
d = 1.836/(0.15*5*9.81)
= 24.9 cm
the block will reach to the slope after 10 cm
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