A Race Between Two Boats
Two boats are having a race on a frictionless horizontal lake. Boat A has a mass of
m
, while boat B has a mass of
2
m
. The boats have identical sails, so the wind exerts the same constant force
F
on each of them.
Which boat has the greater kinetic energy as it crosses the finish line?
Image credit: Bill Wagner
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3 solutions
Discussions for this problem are now closed
This is an interesting problem to me. Most seem to assume that the force on the boats stays constant.
The problem is, with constant wind, the force lessens as the boats go faster. The wind cannot accelerate a boat faster than the wind is pushing. If the boats sail long enough to attain (close to) the same speed as the wind, then the heavier boat will have more kinetic energy!
This is the situation in most boat races I've seen. It's long enough that the boats go nearly the speed of the wind.
Therefore my answer was, "Not enough information."
Given the conditions of the problem, I was probably thinking too practically. Maybe someone can tell me where I've gone wrong? :-)
The problem says "the wind exerts the same constant force on each."
here told wind exerts same force , and force means mass* accelaration so one boat as have more mass than other so acc. of one is less than other so there is difference in velocities of the two at the final point and this differance is again compensated by differance of mass and the net result remains same.
He has made his point.
But I do want to commend your reasoning and critical-thinking on the subject matter.
However, the situation here was an ideal one, one that neglects other factors that are not stated above. I must have overlooked the scenario and forgot to point out that it was an ideal situation, and with that I sincerely apologize.
Well, my thoughts:
F = m a F = M a . a a = > a a = F / M a = > a a = F / m F = M b . a b = > a b = F / M b = > a b = F / 2 m v = u + a t v a = t . F / m v b = t . F / 2 m K E = ½ m v 2 K E a = ½ m ( t F / m ) 2 = ½ m t 2 F 2 K E b = ½ 2 m ( t F / 2 m ) 2 = 4 1 m t 2 F 2 i . e . t w i c e K E a .
Well, I missed the fact that the * distance is same * not time. With that in consideration, answer IS right.
the problem here is that both of them will not reach the finishing line at the same time.
Keep it simple. Work = Force x Distance. The same force was applied over the same distance in both cases. No frictional losses. So, same work in means same energy stored. Has to be the same.
Boat B takes longer to accelerate. So it takes longer to reach the finish line. So the common force exerted on both boats is exerted longer on boat B. Would it not have the greater kinetic every upon reading the finish line?
Ah but energy is equal to work which is force x distance. They go the same distance so kinetic energy is the same.
For Boat A, F=ma......
For Boat B,F=2ma
So, Boat B has half the acceleration of A....
So, using v=u+at for each boat,
V(A)=Ft/m and V(B)=Ft/(2m)
kinetic energy for A= 0.5 m {V(A)}^2
=0.5.{f}^2.{t}^2/m
kinetic energy for B=0.5* 2m* {V(b)}^2
=0.25.{f}^2.{t}^2/m
So, A has greater kinetic energy, isn't it?
But this was not the correct answer, can someone explain why is my process wrong?
I did the same thing (above), I believe the answer is wrong.
Since they start with initial velocity=0 so the total displacement is same not the time they will take is same.So use the third equation of motion
well, doesn't this assume that wind is the only force is acting on the two boats? what kind of a race is that?! Even the picture shows people rowing. So shouldn't we take the rowing force into consideration?! Thats why i picked insufficient data. :-/
Lets say that boats A and B have acceleration x and y respectively no doubt the acceleration of boat of mass M is twice that of boat of mass 2m. Dosent the amount of distance covered matter to determine what velocity the boats obtain? Lets say the acc of boat A is 2 and B is then 4 over the course of 1 second B velocity is 4 and a velocity is 2 . Assume m to be 2 and 1 in which case KE of boat A is 2 2(square)=8 and that of B is 1 4(square)=16 they vary. Then how can they be the same? Can someone tell me what is wrong with my logic or physics in this case?
I just want to know one thing that a=F/m same force is exerting on both boats but masses are different hence acceleration will also be different then how they have same kinetic energy?
as force is same and distance travelled by both the boats are same work done on both the boats by wind is same. but we know if there are dissapative forces work done is equal to change in kinetic energy.as both start from rest initial KE is same ,so is the final.
if mass of body is more then it is necessary that weight of body also more???? plz tell me.
They are displaced by the same amount by the same force. Thus, the work done on each are the same, and their kinetic energies are the same.
It seems reasonable to assume that with sail boats the force exerted on the sails does not all translate as force exerted in the direction of the boat's velocity. If the wind exerts the same force on the sails of both boats, could we not assume the wind velocity is perpendicular to the boat velocity? As the boat speeds up, with no water friction, the sail can be turned to form a greater angle with the wind, maintaining the force of the wind against the sail, but reducing the amount of force being transferred in the direction of the boat velocity. The force exerted on the sail of each ship can be constant, but the effective force in the direction of each boat velocity will not remain constant and will not remain equal for both boats. Maybe I am taking the principles of sailing to literally, or I have misunderstood them.
if sails are constrained then in that case, the letter 'F' in the question would be replaced by 'Fcos(theta)' where theta is (angle between sail track and wind)
They will have the same kinetic energy.
Since the same constant force F acts on both of them, and they are covering the same displacement , say, s , by the Work-Energy Theorem:
W = F s = Δ K = K f − K i
This means that if both boats started from rest such that v i = 0 , then they should have the same final kinetic energy, irregardless of their mass.
Note, also, that the final velocity of boat A is actually greater than that of boat B, to compensate for its lack of mass. That is:
2 1 m A v f , A 2 = 2 1 m B v f , B 2 \ v f , A 2 = 2 ⋅ v f , B 2 v f , A = 2 ⋅ v f , B