A 120 kg box is being pulled up an incline by a rope...

A 12.0 kg box is being pulled up an incline by a rope parallel to the incline that makes an angle of 24.0° above horizontal. There is a coefficient of friction between the box and incline of 0.18. If the box slows down uniformly from 2.2 m/s to 1.40 m/s as it moves a distance of 2.50 m up the incline, what is the force of tension exerted on the box by the rope?

Transcript

00:01 Hello guys in this problem we have a rough inclined plane with angle of inclination 24 degrees we have an object with mass 12 kilogram being pulled with the help of a rope such that the tension in the rope is t the speed of the box reduces from the initial value v1 of 2 .2 meters per second to the final value v2 of 1 .4 meters per second for the displacement along the plane of 2 .5 meter if the coefficient of kinetic friction between the block and the inclined plane is 0 .18 we have to determine the tension t for that let's assume uh sorry let's draw the freeboard diagram the other force is acting on the box is the weight amm in the vertical downward direction a normal force n perpendicular to the plane and to oppose the motion a kinetic friction force now let's take the component of eight perpendicular and parallel to the plane the perpendicular component is m g times cost theta as this angle is theta and the parallel component is mg sine mg times sine theta if we assume cotton axis normal to the plane as y -axis and towards the top as x -axis we can write net force along y is c and from this we get the normal force equals m g cost theta hence the kinetic friction force equals mu k times m g g g gos theta we can apply the work energy equation from the work energy equation network 10 equals the change in kinetic energy now the network is worked in by gravity plus work done by the tension force plus worked on by normal force plus worked on by the friction force and it equals the final kinetic energy minus the initial kinetic energy now work done by normal becomes zero as the normal force is perpendicular to the displacement worked done by gravity is minus of m gd d sine theta work done by the tension is t times t worked on by friction is minus of coefficient of kinetic friction m g t cost theta and it equals half times m times the speed v2 square minus v1 square from this we get the tension equals um times v2 square minus v1 square over 2d minus m g times sine theta plus coefficient of kinetic friction cost theta.

03:43 Sorry, here it should be plus...

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