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EIOCilyIleaves rough patch? force of 100 Nt is Assume [ Usen lift an = no friction, kg mass calculate; height of 20 m, {a) The work don The the applied change force...

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EIOCilyIleaves rough patch? force of 100 Nt is Assume [ Usen lift an = no friction, kg mass calculate; height of 20 m, {a) The work don The the applied change force . what Tiafse weight $ potential mass" velocity energy. wnen 20 m? 100 9 Dlece of

EIOCily Ileaves rough patch? force of 100 Nt is Assume [ Usen lift an = no friction, kg mass calculate; height of 20 m, {a) The work don The the applied change force . what Tiafse weight $ potential mass" velocity energy. wnen 20 m? 100 9 Dlece of



Answers

Murimi pushes a 20 -kg mass $10 \mathrm{m}$ across a floor with a horizontal force of $80 \mathrm{N}$. Calculate the amount of work done by Murimi.

Problem 36. And this problem, Force F. Is pulling the mass to KZ along the inclined plane. We have to determine what amount of work will be done by this Force F. To move it from the bottom of this plane to the top of this plane, Because we have to move it by a vertical height of two m. So in previous part, we have calculated the value of force and the value of force was 10 Newton. No, let us see that distance traveled along the inclined path. Let us say this distance is X meter. Therefore, from geometry, we can say that 10 30 degree will be equals two. Sorry, we can use sign 13 straight off then. So sign 30° as equals to two divided by X. Sign 30 is half, that is equals to two divided by X. Therefore, X. S. Equals to four m. So we have got the value of distance along the inclined plane, and we already know the value of force from the part 35, so work then will be forced, multiplied by displacement. So forth is 10 Newton and displacement along the forces. four m, therefore worked and will be Equals to 40 June. So, from the given options option E. is the character option 42.

Hello students in this question we have a letter of mass M equals to 20 kg. And length L equals to 2010 metres which is resting at the angle theta equals to 30 degrees. So we can draw this diagram like this. Okay, so this is the horizontal and vertical wall and here it is the letter. Okay, now this is the slippery Vertical vault. Okay and this is the horizontal axis. So this angle Tita is given this. Okay, so this angle will be 60°. Okay? No, the weight MG will act from here in this vertical downward direction. Okay. And that we have to calculate the force exerted by the letter on the wall. That is animal. Okay, so here due to slippery action, the friction force will act in this direction, supposed have to and this will be the normal force and two. So I suppose this is point A and this is point to be So we can take the talk about point to be this will be equal to zero because the letter is addressed. So the talk of these two forces will be zero. So the talk of this force MG will be MG market LP to mark. Labour costs 60 degree minus off and one more club L sign 60 degree. Okay and that is equal to zero. So from here we get and when it is equal to MG court 60 degrees development. So m is 20 kg that is 20 into 10 to the power three ground and G is 9 80 centimeters per second squared Multiplied by court 60 will be won by under or three multiplied by two. So we get from here. Normal force, that is the force Exerted by the letter on the wall. It is equal to 5.7 per click bait and to the power six, dying. Okay, so this is the answer for this problem. Okay. Thank you.

In this question, we will learn about some basic concept of the work, energy and power. The question is a body of master in KK is moving on a horizontal surface by playing a force of 10 Newton in forward direction. If the body moves with constant velocity, the work done by the applied force for displacement of two m is what we have to find. Right? So in this question we have The mass of the body that is 10 Kg. The force I played on the body is then you're done And the displacement of the body is two m. Now, since the force and displacement are in the same direction. Therefore, the theater. That is the angle between force and displacement. Yes, joe degree. So the work done by the force applied would be equal though the force multiplied by the displacement, multiplied by car. Stay to where do you think is the angle between force and displacement? So we can substitute the values The value of four. System you've done. This placement is two m and caused your degree. So this is equal to 20 tool. Right. So they were done by the For supply this 20 jewel and it is matching with option A. Therefore, is that correct correct solution? Mhm

Question number 70. From the given information, we can write the change in kinetic energy, which is half times moss within bracket initial lost a square minus finally lost a square, plus the workmen by the friction Polls is equal to integration off the given force limit from minus 4 m. Two plus 4 m and force has given us minus five. Eggs is square less seven x the X now integrating this within limits from minus 4 to 4. We will get minus five eggs you upon three less seven X is squired upon to within limits minus four. No, that's for. And here half into Mars It's too giddy into initial lost zero minus nine. It's square plus for gun by the frictional force So this will give us minus 81. Plus we're done by the frictional force and substituting the limits we will get minus 213 0.33 So the were done by the friction is w equals minus 132 0.3. Jules, this world is negative because of friction. Does the negative work


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