Santa's Sleigh (Newton's 2nd Law-1D) Acceleration, Horizontal Motion

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SUMMARY

The discussion centers on calculating the takeoff speed of Santa's sleigh, which has a mass of 62.5 kg and is pulled by eight reindeer exerting a total average force of 8350 N over a distance of 60.0 m. Participants emphasize the application of Newton's 2nd Law (F=ma) and the concepts of work and energy to solve the problem. The ambiguity in the problem statement regarding the nature of the force (constant vs. average) is highlighted, with suggestions to assume a constant force for simplification. The consensus is that the average force refers to the total force exerted by the reindeer.

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Angelx26
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Homework Statement
Santa's sleigh, which has a mass of 62.5 kg (without Santa as an occupant), is pulled by eight tiny reindeer with an
average force of 8350 N over a 60.0-m stretch of snow before becoming airborne. Assuming the sleigh started
from rest, what was the sleigh's takeoff speed? (Ignore friction).

I am using a study guide that doesn't show much work, only shows the answers. I know it involves Newton's 2nd Law.
Relevant Equations
F=ma.
All I know is that it involves Newton's 2nd law F=ma.
Im wondering if I use average acceleration (vf-vi)/time
I'm not sure where else to go from here.
 
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Angelx26 said:
Homework Statement: Santa's sleigh, which has a mass of 62.5 kg (without Santa as an occupant), is pulled by eight tiny reindeer with an
average force of 8350 N over a 60.0-m stretch of snow before becoming airborne. Assuming the sleigh started
from rest, what was the sleigh's takeoff speed? (Ignore friction).

I am using a study guide that doesn't show much work, only shows the answers. I know it involves Newton's 2nd Law.
Homework Equations: F=ma.

All I know is that it involves Newton's 2nd law F=ma.
Im wondering if I use average acceleration (vf-vi)/time
I'm not sure where else to go from here.

Do you think each reindeer pulls with a force of 8350N, or is that the total? It's not clear to me what is meant.

Where is Santa? Why isn't he in his sleigh? That's another puzzle.

And, yes, you can take ##F_{avg} = ma_{avg}##
 
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Since the reindeer pull with an average force, then that implies an average acceleration, to me.
 
One moment while we wait for @haruspex to call attention to the blunder in the problem statement...

The problem setter is begging for you to apply the concepts of work and energy here.
 
scottdave said:
Since the reindeer pull with an average force, then that implies an average acceleration, to me.
Yes, but without knowing the duration that does not help.
jbriggs444 said:
One moment while we wait for @haruspex to call attention to the blunder in the problem statement...

The problem setter is begging for you to apply the concepts of work and energy here.
Thanks for the invitation... you know I can't resist.
@Angelx26 , sadly there is not enough information to answer the question. Had you been told it was a constant force then there would have been several ways to solve it, e.g. using work=force x distance, or F=ma together with v2-u2=2as. The two are essentially the same.

As it is, it depends how the force varies with time. E.g. if no force at all for 9 seconds then 83500N for one second then work = 83500N x 60m = ½mv2, but if a constant force then work is only 8350N x 60m.
 
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I agree that the question should have been stated clearer. My best guess is that by the "average force" of ##8350\text{ N}## it is meant that the total force exerted by the eight tiny reindeer at every moment is ##8350\text{ N}##, i.e. the sleigh is pulled by a constant force of ##8350\text{ N}##. As was discussed, if the force varies with time the given information is incomplete and ##8350\text{ N}## per reindeer is probably too much, even for Santa's reindeer.

So I'd just solve it for the simplest case of a constant force, but yeah, the wording is somewhat murky.
 
I guess the word "average" is often used in these problems because a constant force sounds unrealistic, so the intended assumption is that the force stays close to a certain value, with some relatively small variation. This would support a further simplifying assumption that the force is constant for the purpose of calculation.

A better statement would be to assume that the force is approximately constant.
 
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