A horizontal force F1=40 N and a vertical force F2=30 N are acting on a 5 kg mass.
If the mass is displaced a horizontal distance r=2 m as shown in the figure, what is
the work in J done by the force F2?
F
x
O a 100
O b. 50
OC 80
Od 0
O e 60
Answers
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.
(a) In this case, P=8.0 N leads to
F
N
=(2.5kg)(9.8m/s
2
)–10N=14.5N.
This implies ,f
s,max
=μ
s
F
N
=6.6N, which is larger than the 6.0 N
rightward force. Thus, the block (which was initially at rest) does not move. Putting a = 0
into the first of our equations above yields a static friction force of f=P=6.0 N.
(b) In this case, P=10 N, the normal force is
this implies ,f
s,max
=μ
s
F
N
=5.8N , which is less than the 6.0 N rightward
force – so the block does move. Hence, we are dealing not with static but with kinetic
friction, which Eq.reveals to be f
k
=μ
k
F
N
=3.6 N.
(c) In this last case, P=12 N leads toF
N
=12.5 N and thus to, f
s,max
=μ
s
F
N
=5.0N ,
which (as expected) is less than the 6.0 N rightward force. Thus, the block moves. The
kinetic friction force, then, is f
k
=μ
k
F
N
=3.1N
Explanation:
hope it helps