A 25 kg box is released on a 23° incline and accel...

Question

A 25 kg box is released on a 23° incline and accelerates down the incline at 0.30 m/s2.A) Find the friction force impeding its motion.B) What is the coefficient of kinetic friction?

Patrick Ford · Accepted Answer

Newton's second law of motion:∑F=ma[readmore]The frictional force acts up the incline, while the force acting on the box down the incline is given by mg sinθ.From Newton's second law:mgsinθ - fk = maA)The frictional force is given by:fk=mgsinθ-ma=m(gsinθ-a)=(25)[(9.81)sin(23°)-0.30]fk = 88.33 NThe friction force impeding the motion of the box is 8.33 N. B)The friction force is also expressed as:fk=μkNWe'll consider motion in the y-direction.We know that there is no motion along the y-direction. Therefore, ay = 0∑Fy=maN-mgcosθ=0N=mgcosθFrom the boxed equation for friction force:fk=μkmgcosθμk=fkmgcosθ=88.33(25)(9.81)cos(23°)μk = 0.39The coefficient of kinetic friction is 0.39.

Problem: A 25 kg box is released on a 23° incline and accelerates down the incline at 0.30 m/s2.A) Find the friction force impeding its motion.B) What is the coefficient of kinetic friction?

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Problem Details

Problem: A 25 kg box is released on a 23° incline and accelerates down the incline at 0.30 m/s2.A) Find the friction force impeding its motion.B) What is the coefficient of kinetic friction?

FREE Expert Solution

Problem Details

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