if-it-takes-45-mathrmn-of-force-to-lift-a-box-38-mathrmm-how-much-work-is-done-to-the-box

Name: if it takes 45 mathrmn of force to lift a box 38 mathrmm how much work is done to the box
Uploaded: 2020-10-19
Duration: 1 min 19 s
Channel: Numerade
Description: If it takes 45 $\mathrm{N}$ of force to lift a box $3.8 \mathrm{m},$ how much work is done to the box?

Question

If it takes 45 $\mathrm{N}$ of force to lift a box $3.8 \mathrm{m},$ how much work is done to the box?

David Collins · Accepted Answer

This next question, similar to the previous question, asks you to calculate the amount of work that is being done by a particular process. The amount of work that is being done by the particular process can be calculated using one of two equations that are available in the textbook. One equation we used earlier that was work is equal to pressure multiplied by a change in volume. However, we don&#x27;t have an expanding gas. In this problem, we simply no a force that is being applied over a distance. So the more appropriate equation is work is going to be equal to this force in Newtons, multiplied by the distance in meters. The force that is being applied is 45 Newtons, and the distance over which this force is being applied is 3.8 m. So when multiplied together we get 171 Newton meters, which is the same thing as 171 jewels. One Newton meter is a Jule

Ankit Gupta · Answer

world in W is equal toe effect er dot product with a B veteran that is equal to 60 multiplied by 20 multiplied by cause off 38 degree minus 12 dignity that is equal to 1200 multiplied by course, off 26 degree that is approximately equal to 1079 foot pound. It means work Island is equal to 1079. £4.

Averell Hause · Answer

here. The work done would be equal to the force. Times of Displacement Times Co sign of Fada, the force of 628 Newtons. This would be for the force applied to the rope multiplied by 15 0.0 meters, multiplied by co sign of 46 0.0 degrees. And we find that then the work is equaling 6.54 times, 10 to the third jewels. And this would be our final answer. That is the end of the solution. Thank you for watching.

Eric Mockensturm · Answer

So we&#x27;re told we&#x27;re pushing a 67 cala graham box across the floor where the kid pneumatic coefficient of friction is 0.55 And we&#x27;re pushing it in the velocity of 05 m per second. And they ask us number one, how much power is needed to push the box um And how much work do you do push the box for 35 seconds? So what we need to know is the the force that we&#x27;re pushing a consuming, we&#x27;re pushing at a constant velocity that this force that we&#x27;re pushing at is equal to the friction force that&#x27;s resisting are pushing. So that means then that friction forces the static coefficient or the cinematic coefficient of friction times the normal force. The normal force is just the weight and the weight is just MG. So the power equals F times V. Which is an F S M U K MG. So we have U. K. Mg. Times V. Plugging all these values in here, we get that that is 0.18 kilowatts. So that&#x27;s how much power were exerting pushing this box. And if we push it at this this constant rate, the constant force, constant velocity then you know p is a constant, not a function of time. So we can just take the work we did is p times the change in time or the time over which that work was acting. So we get 018 times 35 seconds. And that winds up being um um 6.3 killers rules. So that&#x27;s how much work we did and pushing this along.

Averell Hause · Answer

the work done. W would be equal to the force times the distance temps co sign of data, the angle between the force and the distance traveled, or the displacement. And so this would be equaling 22 because each person exerts a force of 225 Newton. So two times to 100 and 25 Newtons times 15 meters. And then the angle between the force and the displacement we could say co sign of 15 degrees, and we find that then the work is equaling approximately 6.5 times, 10 to the third jewels. That is the end of the solution. Thank you for watching.

Lili Schantz · Answer

for this problem. We want to find how much work is done pulling the box 25 feet along the ramp that is inclined at 10 degrees from the horizontal with a rope that forms an angle of 33 degrees to the horizontal and a force of £80. So we first seem to find Earth data because we know that work equals, um the magnitude of force times the magnitude of our distance. Times co sign data in the data is gonna be our angle between vectors. So we have this 10 degrees of our incline, and then we have the 33 degrees of the rope. So to find data, we want to find this angle here. And to do that, we need to do 33 minus 10 which is 23 degrees. So our data is 23. So now we can just plug everything and so we have work is 80 times 25 times co sign 23 degrees in that equals 1841 foot pounds.

Problem

Research and describe three ways that a rocket en…

Problem 4 Easy Difficulty

If it takes 45 $\mathrm{N}$ of force to lift a box $3.8 \mathrm{m},$ how much work is done to the box?

Answer

171 $\mathrm{J}$

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