Lili pulls a 600 kg crate across a horizontal floor for a...

Lili pulls a 6.00-kg crate across a horizontal floor for a distance of 3.00 m as shown. The tension force in the rope is 86 N and the angle is 30.0 degrees with respect to the horizontal. The frictional force on the crate is 12.0 N. Assuming the system is crate+earth+floor, match the answers with questions. What is the change in gravitational potential energy (ΔUg)? A. 223 J What is the change in thermal energy (ΔEth)? B. -259 J What is the work done by friction? C. 36 J What is the work done by you? D. -36 J E. -187 J F. -223 J G. none of the given H. 0 J I. 259 J

Transcript

00:01 Hello everyone here the question consists three parts part one part two and part three so in part one we have to find out the work done that can be obtained as the area of the curve of the force versus position guard and this is graph is given so now for part one from the graph the work done can be obtained as the area of the curve of the force versus position graph so now we can retain as from the graph w equals to half multiplied by 4 meter multiplied by 10 newton plus 3 meter multiplied by 10 newton plus half multiplied by 5 meter multiplied by 10 newton.

00:49 So further solving this we will get the value of work done.

00:53 W equals to 75 june.

00:57 So this is our final answer for part 1.

01:01 So part 2 we can draw the diagram according to question so here we have to find out the ferocity so now let's see the answer so work done by tension force of string we can retain as from the figure w t equals to t vector multiplied by b vector so from here you can retain as t b cos of theta since theta is the angle between tension and the distance which is equal to 30 degrees so from here you can retain as 600 multiplied by 10 multiplied by cost of 30 so further solving this you will got w t equals to 3000 root 3 jule so work done by dynamic friction we can return as w suffix f x k a equals to fictional force f k multiplied by distance d so here angle between f k and distance is 180 degree so further solving this we can return as f k the cause of 180 degree so further solving this here we will got w suffix f k equals to minus f k b where we know that f k equals to mu k multiplied by normal force n so from the figure we can see that n plus t signed of 30 equals to m g therefore from here we will got the value of n equal to 50 g minus t sine of 30 so further solving this we will got the value of n equals to 190 newton so therefore we will got fictional force f k equals to mu k value is already given which is 0 .6 multiplied by 190 newton so after solving this we will got fictional force f k equals to 114 newton so therefore using the above equation work done against fictional force f k equals to you can retain as 114 newton multiplied by distance is 10.

03:59 So from here we will got w suffix fk equals to minus 1140 june...

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