How is acceleration impacted when mass remains the same but the amount of force is decreased?

1 answer:

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The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

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A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (

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Answer:

At the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).

This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).

If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.

Explanation:

From the question,

The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).

Let the truck's momentum be P and the truck's speed be v,

Then we can write that

P∝v

Then,

P = kv

Where k is the proportionality constant

From the question,

At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,

To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is

50600 ÷ 23 = 2200

Hence, at the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).

Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then

This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).

From

P = kv

Then, k = P/v

At a moment, P = 50600 kg · m/s and v = 23 m/s

∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg

k = 2200 kg

To determine the truck's momentum if the truck is traveling at 16 m/s

From