Question

Use the mass of the block and the angle of incline to determine how much of the weight of the block is acting down the plane. This weight must be overcome by the force pulling the block up the incline. The rest of the force pulling the block up the incline is overcoming friction . Calculate this force overcoming friction and calculate how much work is lost in overcoming friction. mass = 0.745 kg and the angle is 45 degrees.

Answer 1

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Forces acting on the body are component of gravitational force $mg\sin{\theta}$ down the plane, component of gravitational force $mg\cos{\theta}$ perpendicular to the inclined plane towards it, Normal reaction perpendicular to the inclined plane away from it, frictional force $F_{f}$ down the plane and force applied by person up the plane.

Net force perpendicular to the plane is zero, so $N=mg\cos{\theta}$

when the applied force just overcomes friction and component of gravitational force down the plane, net force parallel to the plane is zero.

$F=mg\sin{\theta}+F_{f}$

since the body is sliding upwards, frictional force is $F_{f}=\mu_{k}N=\mu_{k}mg\cos{\theta}$

Hence force necessary to overcome friction and gravitational force is $F=mg\sin{\theta}+\mu_{k}mg\cos{\theta}=(5.17)(1+\mu_{k})\,N$

If the displacement iof the body is up the plane, work done by this force is $FS=(mg\sin{\theta}+\mu_{k}mg\cos{\theta})S$

Of this work done to overcome friction is $(\mu_{k}mg\cos{\theta})S=(5.17\mu_{k}S)\,J$