A 185N sled is pulled a distance of 180m. The task requires 5200J of work by frictioin and is done by pulling a rope at an what angle? Coefficient of friction 0.30
PLEASE SOLVE :(
PLEASE SOLVE :(
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I'm going to assume that the sled was being pulled at a constant velocity:
So I did my best on this, I'll walk you through my steps so that you can make sure that I'm not wrong:
Knowing that work=force*distance, the work done by friction must be:
work=forcefriction*distance
5200=ff*180
ff=28.89N
knowing that the coefficientoffriction= Ff / Normalforce:
.3 = 28.89 / N
Normal Force = 96.27N
now since the sled is being pulled at an angle, there are upward and horizontal components to the force pulling the sled. The normal force deals with the net vertical forces on he sled, so:
NormalForce=mg-VerticalForce
96.27 = 185N - Fv
Fv = 88.7N
Since I assumed the sled was moving at a constant velocity, that would mean the force of friction is equal to the horizontal force. we know the force of friction is 28.89N, so that is also the horizontal force.
Consider these forces to be a right triangle with the pulling force as the hypotenuse and the vertical and horizontal components as legs. Do some basic trig to find theta:
Tan(theta)=opp/adj
Tan(theta)-Fv/Fh
Tan(theta)=88.7 / 28.89
tan^-1(88.7 / 28.89) = theta
theta = 71.96degrees
that's my best attempt based on the given information, I hope that I at least got you on a train of thought if my answer was incorrect, but I'm pretty confident that it's right because there's not much else you can do given the information. Good luck!
So I did my best on this, I'll walk you through my steps so that you can make sure that I'm not wrong:
Knowing that work=force*distance, the work done by friction must be:
work=forcefriction*distance
5200=ff*180
ff=28.89N
knowing that the coefficientoffriction= Ff / Normalforce:
.3 = 28.89 / N
Normal Force = 96.27N
now since the sled is being pulled at an angle, there are upward and horizontal components to the force pulling the sled. The normal force deals with the net vertical forces on he sled, so:
NormalForce=mg-VerticalForce
96.27 = 185N - Fv
Fv = 88.7N
Since I assumed the sled was moving at a constant velocity, that would mean the force of friction is equal to the horizontal force. we know the force of friction is 28.89N, so that is also the horizontal force.
Consider these forces to be a right triangle with the pulling force as the hypotenuse and the vertical and horizontal components as legs. Do some basic trig to find theta:
Tan(theta)=opp/adj
Tan(theta)-Fv/Fh
Tan(theta)=88.7 / 28.89
tan^-1(88.7 / 28.89) = theta
theta = 71.96degrees
that's my best attempt based on the given information, I hope that I at least got you on a train of thought if my answer was incorrect, but I'm pretty confident that it's right because there's not much else you can do given the information. Good luck!