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A box with a mass of 0.5 slugs lies on an inclined plane making an angle of 30o with the horizontal. If the coefficient of kinetic friction between the box and the plane is 0.4, what is the magnitude of the force that must be applied parallel to the incline to keep the box moving down the incline at a constant speed

User Kiang Teng
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1 Answer

2 votes

Answer: 10.98 N

Step-by-step explanation:

Given

mass of box is
m=0.5\ slugs\approx 7.3\ kg

The coefficient of kinetic friction is
\mu= 0.4

It is given that on the application of force box started moving with constant speed i.e. there is no net external force


\Rightarrow F+mg\sin 30^(\circ)=\mu mg\cos 30^(\circ)\\\Rightarrow F=mg(\mu \cos 30^(\circ)-\sin 30^(\circ))\\\Rightarrow F=7.3* 9.8(0.866*0 .4-0.5)=-10.98\ N\\

The negative sign indicates direction of force is opposite i.e. it must be applied upwards

User SuperCow
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