牛顿第二定律实验计算求助：橡皮筋弹射物体实验解析

Hey there, let’s walk through this experiment and clear up those Newton’s second law misconceptions you’re dealing with—this is a common spot folks get tripped up, so let’s break it down step by step.

First, a critical note that’s probably part of your confusion: that 5N from the rubber band isn’t a constant force. Rubber bands exert a variable force (it drops as the band relaxes), so we can’t treat it like a steady 5N pushing the object the entire time. Instead, we’ll use kinematics and work-energy principles since we have displacement and time data to work with.

We’ll start with reasonable simplifications for short, slow motion like this:

• The rubber band transfers all its energy/impulse to the object instantly when the string is burned (so we only need to analyze the motion phase where friction and air resistance are the only forces acting)
• The combined force of friction and air resistance is constant (this holds well for low-speed, short-distance movement)

Since we’re assuming constant deceleration (from constant resistance), average velocity is straightforward:
v_avg = Δx / Δt = 0.5m / 0.5s = 1 m/s

For constant acceleration/deceleration, average velocity is also the average of the initial velocity (right after the rubber band pushes the object, v₀) and final velocity (which is 0, since the object stops):
v_avg = (v₀ + v_f) / 2
Plugging in v_f = 0:
1 m/s = (v₀ + 0) / 2 → v₀ = 2 m/s

Using the kinematic equation for velocity over time:
v_f = v₀ + aΔt
We know v_f = 0, v₀ = 2 m/s, and Δt = 0.5s, so solving for a:
0 = 2 + a(0.5) → a = -4 m/s²
The negative sign just means the acceleration is opposite to the object’s motion (it’s decelerating).

From Newton’s second law (F_net = ma), the only net force acting during the object’s motion is the resistance force f (opposing motion, so it’s the negative of the net force if we take motion direction as positive):
-f = ma
Rearranged to solve for f:
f = -ma

Wait a quick gap here: we don’t know the object’s mass m! If you can measure the mass with a scale, you can plug it in directly. For example, if the object weighs 0.25 kg:
f = -(0.25 kg)(-4 m/s²) = 1 N

To double-check, the work done by resistance should equal the initial kinetic energy of the object (since it comes to a complete stop):
-fΔx = -0.5mv₀²
Using our example mass of 0.25 kg:
-f(0.5) = -0.5(0.25)(2)² → -0.5f = -0.5 → f = 1 N
Perfect, this matches our earlier result!

The big mistake many people make here is treating the rubber band’s 5N as a constant force for F=ma. Remember:

• Newton’s second law applies to the net force at any instant. Since the rubber band’s force changes as it relaxes, we can’t use F=ma directly with 5N unless we have data on how the force varies over distance or time. Impulse or work-energy are far better tools for variable forces like this.
• That 5N is almost certainly the maximum force the rubber band exerts (when fully stretched), not a steady force throughout the push. That’s a super common mix-up!

If you can measure the object’s mass, you’ll get the exact resistance force. Let me know if you need to adjust any assumptions or have more data to work with!

内容的提问来源于stack exchange，提问作者RMM