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Jac8897
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A boy pulls a 14.6kg sled 12.5m at a constant speed across a horizontal floor. if the rope makes a 32.4degrees?
angle with the horizontal and the coefficient of friction is .489, what is the work done by each of the forces acting on the sled?
2 days ago - 2 days left to answer.
Additional Details
my work,
from the free body diagram I get
EFy= n+Fsin@-mg=0 "@ = theta"
EFx= Fcos@-f=0 "constant speed"
solve Fy= n=mg-Fsin@
plug it into Fx
friction=un
Fcos@-u(mg-Fsin@)=0
Fcos@-umg+uFsin@=0 >>
Fcos@+uFsin@=umg >>
F(cos@+usin@)=umg>>
F=(umg)/(cos@+usin@)
F=63.3N >>
work of the force
WF=63.3N(12.5)cos0>> WF= 791J
no solve by friction
Fcos@-f=0 >>> f=Fcos@
f=(63.3N)cos(32.5) >> f=53.4N
work of friction
Wf=(53.4N)(12.5M)cos180 >> Wf= -668J
I ignore the N force and the gravity and Fsin@ because they are perpendicular.
is my physics process good?
is my answer right?
thanks
angle with the horizontal and the coefficient of friction is .489, what is the work done by each of the forces acting on the sled?
2 days ago - 2 days left to answer.
Additional Details
my work,
from the free body diagram I get
EFy= n+Fsin@-mg=0 "@ = theta"
EFx= Fcos@-f=0 "constant speed"
solve Fy= n=mg-Fsin@
plug it into Fx
friction=un
Fcos@-u(mg-Fsin@)=0
Fcos@-umg+uFsin@=0 >>
Fcos@+uFsin@=umg >>
F(cos@+usin@)=umg>>
F=(umg)/(cos@+usin@)
F=63.3N >>
work of the force
WF=63.3N(12.5)cos0>> WF= 791J
no solve by friction
Fcos@-f=0 >>> f=Fcos@
f=(63.3N)cos(32.5) >> f=53.4N
work of friction
Wf=(53.4N)(12.5M)cos180 >> Wf= -668J
I ignore the N force and the gravity and Fsin@ because they are perpendicular.
is my physics process good?
is my answer right?
thanks