(adsbygoogle = window.adsbygoogle || []).push({}); 1. A 2.00 x10^2-N force is pulling an 75.0-kg refrigerator across a horizontal surface. The force acts at an angle of 23.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 9.00 m. Find (a) the work done by the pulling force, and(b) the work done by the kinetic frictional force.

I need help with part b of this problem

mass: 75 kg

distance: 9 m

Force: 200

theta: 23 degrees

2. Work= Force*distance* Cos(theta)

Friction=mu*Normal Force

3. (mass*9.8)-Tension*Sin(theta)=Normal Force

((mass*9.8)-tension*Sin(theta))*mu=Friction Force

Friction Force*distance=work done by frictional force

(75*9.8) - (200*sin(23)) = Fn

735-78.146= Fn

656.85377 = Fn

so,

656.85377*.2=Friction force

131.371= Friction force

so,

131.371*9=work

1182.337 J = work

I have a few questions:

First, what am I doing wrong?

Is cos(theata) = cos(0) = 1

and

would the final answer for work be negative because this is the work of friction?

Any help would be appreciated greatly!

Thank you ahead of time :)

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