A force F of magnitude 20 N is applied to a block of mass 2 kg that lies on a rough, horizontal surface as shown in figure below. The coefficient of kinetic friction between the block and surface is 0.4. What is the magnitude of the acceleration of the block?
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Answered by
0
Answer:
In addition to the forces already shown in Fig. , a free-body diagram would
include an upward normal force
F
N
→
exerted by the floor on the block, a downward m
g
→
representing the gravitational pull exerted by Earth, and an assumed-leftward
f
→
for the
kinetic or static friction. We choose +x rightward and +y upward. We apply Newton’s
second law to these axes:
F−f=ma
P+F
n
−mg=0
where F=6.0 N and m=2.5 kg is the mass of the block.
Answered by
1
Answer:
Fcosθ - µk (mg - Fsinθ) == ma
20N (cos53°) - 0.4 (2kg(10m/s2) - 20N(sin53°)) =2kg (a)
10.4N = 2kg(a)
We divide both by 2, so the answer will be:
a = 5.2 m/s2
Explanation:
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