An object with a mass of 6 kg lies on a horizontal surface with friction with a

Question

An object with a mass of 6 kg lies on a horizontal surface with friction with a coefficient of kinetic friction of 0.65. Two additional forces are applied to it. One force of 25 N directed at an angle of 40 degrees below the horizontal and another force in the +x direction. Static friction has been overcome and the mass slides 8 meters in the +x direction with these forces applied to it. After it has slide 8 meters, the net work done by all the forces acting on it is 24 Joules. What is the amount of the second force in Newtons?

Explanation / Answer

Normal force= mg= 6*9.8= 58.8 N

Friction in x direction= mu*N = 0.65*58.8=38.22N

Work done = net force* distance= net force*8 = 24J

Net force in x direction= 3 N

3N = 25*cos(40°) + Fx - 38.22

Fx= 3+38.22 - 25*0.766 N

Fx= 41.22 - 19.15

Fx= 22.05 N

Another Force applied in x direction is 22.05 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.