A 2.00 kg box is at rest on a level, frictionless surface....

A 2.00 kg box is at rest on a level, frictionless surface. The box is pushed for 18.0 m by a completely horizontal force of 40.0 N on the surface. (a) Draw a picture of the situation (making sure to define the system). (b) Draw energy bar diagrams for the box before and after the push. (c) Determine the final speed of the box. (d) If the box had been pushed by a force that pointed down by an angle of 30.0° below the horizontal instead of perfectly horizontally, what would be the final speed of the box? A book (m = 1.0 kg) slides across a level, carpeted floor at an initial speed of 2.50 m/s and comes to rest after 2.80 m. (a) Draw a picture of the situation (making sure to define the system). (b) Draw energy bar diagrams for the box initially (when the box has its initial speed) and finally (when the box has come to a complete stop). (c) Calculate how much kinetic energy was lost. (d) Use the general form of conservation of energy and your answer to the last part to determine the coefficient of kinetic friction between the book and the carpet.

Transcript

00:01 Hello everyone we are going to understand this question here given in the question given mass of log must let it is represented by m which is given 2 .0 kilogram and applied force f is equal to 40 newton and distance post or we can write post distance d is equal to 80 meter so in the part a we have to find the energy so we know that energy sorry in the part a we have to draw the diagram for this question so let this is the block is having mass that is 2kg which is working vertically downward weight is working vertically downward and this is the normal reaction and suppose someone pushing this block with force 40 newton this will be the block diagram for this question when man is crossing this block in horizontal direction that after some time it will reach at this place at this place so here distance between this point and this point it is given 18 meter now in the part v we have to find the energy we know that energy is equal to work done energy is equal to work done now here applied force to displacement into angle between displacement and force now substituting the value in this so we can write 40 into 18 into cos 0 so after calculation we will get 7 120 joules now we have to find the velocity of the object now we know that half mv square is equal to now applying the conservation let the velocity be v let the velocity be so applying the conservation of energy applying conservation of energy applying conservation of energy conservation of energy so we can write half m v square is equal to work done w so substituting the value in this so v square is equal to 2 w upon m again we can write v is equal to square root of 2 w upon m now subtracting the value in this, so we can write 2 into value of w that is 720 upon mass of object that is 2 kilogram.

04:38 So after calculation we will get 26 .83 meter per second hence the required speed of object hence the required speed of object speed of object be 26 .83 meter per second.

05:14 Now solving for point d.

05:19 If force is at angle 30, if force is at 30 degree with horizontal, means if you will draw the diagram, then force will be along to this.

05:45 This is the third theta.

05:49 So here we know that work done, w is equal to f into the cos theta.

05:56 Here it is below axis so it will be minus cos theta.

06:00 It means minus cos minus theta.

06:03 Again, substituting the vary in this...

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