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Physics question acceleration

Please can i have help on this question? I thought the answer is c because the horizontal force is constant and air resistance is negligible and i thought the fact that the fuel is burning wont affect the horizontal force because it will only affect weight which is not parallel so it wont affect the resultant force however the answer is b?

Here is the question; https://app.gemoo.com/share/image-annotation/587376150518231040?codeId=v6gx5EVQYa52N&origin=imageurlgenerator

Thanks!
Reply 1
Imagine you've got this rocket-powered truck on horizontal tracks, right? It's cruising along, and the force pushing it is a steady buddy—doesn't waver. Air resistance? Nah, that's like a gentle breeze to this truck. Now, here's the kicker: the truck's mass is basically the same as the fuel it's guzzling.
[start]So, they light up the fuel, and it's all systems go. The force pushing it is rock-solid, but here's the twist. As the truck burns through fuel, it's shedding weight. And you know what happens when the weight drops? Acceleration cranks up.[/start]
[start]It's like your friend giving you a push on a skateboard. You start with a backpack full of books (that's your initial mass), and your friend's push (the force) gets you rolling. Now, picture ditching some books along the way. Your friend's push is the same, but you're lighter, so you start zooming faster. That's what's happening with the truck. The force stays chill, but as it burns fuel and sheds weight, it's picking up speed.[/start]
[start]So, the deal is, as the fuel disappears, the truck's acceleration steps on the gas. That's why the answer is B[/start]
Reply 2
Original post by RUDMA
Imagine you've got this rocket-powered truck on horizontal tracks, right? It's cruising along, and the force pushing it is a steady buddy—doesn't waver. Air resistance? Nah, that's like a gentle breeze to this truck. Now, here's the kicker: the truck's mass is basically the same as the fuel it's guzzling.
[start]So, they light up the fuel, and it's all systems go. The force pushing it is rock-solid, but here's the twist. As the truck burns through fuel, it's shedding weight. And you know what happens when the weight drops? Acceleration cranks up.[/start]
[start]It's like your friend giving you a push on a skateboard. You start with a backpack full of books (that's your initial mass), and your friend's push (the force) gets you rolling. Now, picture ditching some books along the way. Your friend's push is the same, but you're lighter, so you start zooming faster. That's what's happening with the truck. The force stays chill, but as it burns fuel and sheds weight, it's picking up speed.[/start]
[start]So, the deal is, as the fuel disappears, the truck's acceleration steps on the gas. That's why the answer is B[/start]

So would the horizontal force not matter then, the acceleration will depend on the weight?
Reply 3
Original post by anonymous294
So would the horizontal force not matter then, the acceleration will depend on the weight?

Its just newton 2 so
f = ma
force is constant so acceleration is inversely proportional to mass, so
acceleration = force/mass
As mass (x) decreses, acceleration (y) increases like
https://www.desmos.com/calculator/t4eynssasg
(edited 5 months ago)
Reply 4
Original post by mqb2766
Its just newton 2 so
f = ma
force is constant so acceleration is inversely proportional to mass, so
acceleration = force/mass
As mass (x) decreses, acceleration (y) increases like
https://www.desmos.com/calculator/t4eynssasg

Ohhh ok i was getting confused with the force but I get it now, thank you!!
Original post by anonymous294
Ohhh ok i was getting confused with the force but I get it now, thank you!!

I hope I am not throwing a wet blanket on your acknowledgement of your understanding of the “explanation”, as it seems that you are “coaxed” to believe the following quote is good without questioning:
Its just newton 2nd law so
F = ma

This is simply an abuse of F = ma. We can only apply F = ma for the constant mass problems. If you have an A-level Physics text, refer to it to see how they derive F = ma or refer to the following links.

If F = ma is used in a maths forum to solve variable mass problems, I “think” I can accept the abuse of maths.

Note that especially in physics, we can use the wrong concept(s) to arrive at the correct answer(s). This thread has a post that demonstrates it. 😉

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