2 blocks and a spring, SUVAT and F=ma

In summary: Therefore, the total force applied by the spring, including the force on the stationary truck, is 100 N.
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Homework Statement
Good evening,
This question is significantly simpler than most in here, so I apologise for that!

Truck A and truck B are at rest and connected by a pin with a spring in between them. Truck A has a mass of 20kg and truck B has a mass of 10kg.

The pin is removed and the spring expands applying force to the trucks for 0.6s before falling to the ground. Truck B has a velocity of 3m/s. Truck A also rolls away.

What is the magnitude of the force applied by the spring?
Relevant Equations
v=u+at
F=ma
I have calculated the acceleration of truck B from v=u+at as 5ms/s. The force applied to truck B is therefore 5x10=50N.

I am unsure whether this question is poorly worded, but I feel a reasonable assumption is that the force applied to truck A would be the same as truck B, without knowing its final velocity but only knowing that it moved. I therefore believed that the total force applied by the spring was 100N.

I have been told that the answer is 50N but cannot understand why. I don't understand how if a spring can exert 50N of force on truck B it can move truck A without any additional force. I didn't know whether this was something to do with Newton's third law? However this q seems to be converting coiled spring energy to kinetic energy, something we have not yet studied.

I apologise in advance for probably butchering various physics terms, having only recently started studying the subject. I have tried to search the Internet for advice but keep getting references to hookes law which I have not yet studied and have been told is not required...

Thank you very much in advance for helping!
 
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  • #2
Ollz said:
What is the magnitude of the force applied by the spring?I therefore believed that the total force applied by the spring was 100N.
First, there is a flaw in the question. Springs do not generally apply a constant force as they expand, so it should either specify that it does or ask for the average magnitude of the force. Fortunately, in v=u+at, a is the average acceleration.

Adding the forces exerted by opposite ends of a spring, or of a string in tension etc., is a common misunderstanding.
Strictly speaking, it makes no sense to ask what force something exerts unless we specify what body it is exerting the force on. If the spring has mass and is accelerating it will exert a different force at each end. Moreover, the forces are in opposite directions, so if we add them up they will at least partly cancel.
If we ignore the mass of the spring, it necessarily exerts the same force at each end, so the question is ok because it doesn’t matter whether it is asking about A or B since the forces will be the same. So it is asking what force it is exerting on each,
 
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  • #3
I agree that the question is asking for the average force because the force exerted on either mass varies with time. If that's the case, there is no need for Hooke's law or energy considerations. There is enough information to find the average force using the impulse equation, ##\Delta p=\bar F \Delta t##. The SUVAT equation for the average velocity also works if you start from it and multiply both sides by the mass.
 
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Thank you very much haruspex and kuruman. What a wonderful resource this forum is!
 
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  • #5
Welcome @Ollz !

Just imagine that both trucks are not moving at all after the spring is released (both trucks are stuck in mud, for example).
You have calculated the force that the spring applies in one direction (pushing truck B).

In order to do that, the spring must have some support at the other end.
That support is “feeling” exactly the same magnitude of force (50 N), only that this force points in a direction that is diamelterly opposed.
Those two forces must balance or cancel each other, otherwise the spring would be accelerating.

For a second example, replace the spring with a steel bar.
Now, back each truck onto that bar at the same time.
One truck is pushing back with 50 N of force.
No more, no less than 50 N will be transferred through the bar onto the back of the other truck.
If that second truck wants to stay put, it must push back with 50 N of force.
 
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