1 (A) A 2 kg block is kicked up a 37 degree incline with an initial speed of 4m/
ID: 1956868 • Letter: 1
Question
1 (A) A 2 kg block is kicked up a 37 degree incline with an initial speed of 4m/s. If the coefficient of kinetic friction between the block and the incline is 0.2, determine how far the block travels before stopping and sliding back down.
(B) Once back at the bottom, the same 2kg block is now on a horizontal surface with a coefficient of static friction of 0.3. Determine the minimum force F that must be applied to the block at an angle of 53 degrees below the horizontal in order to move the block.
Not sure how to begin either parts.
Thanks in advance.
Explanation / Answer
First we would find the value of Normal Reaction
W=mg
=(2)(9.81)
=19.62N
sin37=RN/19.62
RN=19.62sin37
RN=11.81N
FR=RN
FR=(0.2)(11.81)
=2.362N
FD=ma
=(2)(-9.81)
=-19.62N
ma=FD - FR
2a=-19.62 - 2.362
2a=-21.982
a=-10.99
2as=v2 - u2
2(-10.99)s=-(4)2
s=0.728m
The minimum force required to move the Block should be greater than the Friction Force
RN=W + Fsin53
RN=(2)(9.81) + Fsin53
RN=19.62 + 0.799F---1
FR=RN
Fcos53=(0.3)RN
0.602F=(0.3)RN
F=0.498RN---2
F=0.498(19.62 + 0.799F)
F=9.777 + 0.398F
F - 0.398F=9.777
0.602F=9.777
F=16.24N
FR=Fcos53
=16.24(0.602)
=9.78N
Therefore the minimum force required is 10N
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