A 5.51 box sits at rest at the bottom of a ramp that is 7.88 long and that is in

Question

A 5.51 box sits at rest at the bottom of a ramp that is 7.88 long and that is inclined at 35.0 above the horizontal. The coefficient of kinetic friction is 0.40, and the coefficient of static friction is 0.50. What constant force , applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 4.26 ?

Explanation / Answer

To find the net force acting on the object we have to find the acceleration on it so we use the equation s=0.5*a*t^2 where s is the distance along the incline and t t is the time of travel a is acceleration we get ==>a=0.868434 m/sec^2 let the force applied be F resolving the forces along the incline and perpendicular to it we have normal reaction N=mgcos(35) F-mgsin(35)-Fr=m*a Fr is the frictional force on the block Fr=mu*normal reaction hence we have F-mgsin(35)-mu*mgcos(35)=m*a substituting the values we get ==>F=57.87 newtons

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