Slanting surfaces with friction a 50 kg box slides on the...

Slanting Surfaces With Friction: A 5.0-kg box slides on the surface on a ramp that rises at 37° above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is 0.60. a)What is the magnitude of the acceleration of the box if it is sliding up the ramp? b)What is the direction of the acceleration of the box if it is sliding up the ramp? c)What is the magnitude of the acceleration of the box if it is sliding down the ramp? d)What is the direction of the acceleration of the box if it is sliding down the ramp?

Transcript

0:00 Hi.

00:01 Here in this given problem first of all this is a ramp which is inclined at an angle of 37 degree above the horizontal over it there is a box which is sliding either up along the incline or down along the incline.

00:23 Mass of this box if it is m its weight will be acting vertically downward which may be resolved into two components one of the component normal to this inclined plane making same angle as 37 degree so its component this component will be having a value m g cost 37 degree and another component along the inclined plane that is m g sign 37 degree then normal reaction exerted by this ramp over this box and that will be equal to mg cos 37 degree.

01:05 Here, mass of the box is given as 5 .0 kilogram.

01:12 This angle theta already we have seen it to be 37 degree coefficient of kinetic friction.

01:19 This is 0 .60.

01:21 Now in the first part of the problem, if the box is sliding up then obviously the force of friction will be acting downward.

01:39 So the force of friction kinetic friction downward along the incline means it will be acting like this.

02:08 And the box is moving up.

02:10 So net force acting means opposing the motion.

02:15 So the net force which will be opposing the upward motion of this box that will be given as f is equal to one of them is m g sign 37 degree component of the weight downward and force of friction which is given as mu k times the normal reaction and we know normal reaction is m g cost 37 degree and as the force is opposing in nature so we will take a negative sign here then using newton's second law motion the force may be written as the product of mass of the box with its acceleration so here it will be m is equal to and we can take this m as a common out here leaving behind 9 .8 into sine 37 degree plus mu k which is 0 .60 times again 9 .8 into cost 37 degree canceling this m finally acceleration here comes out to be equal to minus 10 .6 meter per second square acceleration in the motion of the box when it is moving upward.

03:39 Answer for the first part of the problem.

03:41 Then in the second part of the problem, we have to find direction of this acceleration...

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