A lumberjack (mass = 103 kg) is standing at rest on one end of a floating log (mass = 255 kg) that is also at rest. The lumberja
1 answer:
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<u>Answer:</u>
2N/cm
<u>Step-by-step explanation:</u>
According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:
where, is the force which is stretching or compressing the spring,
is the spring constant; and
is the distance the spring is stretched.
Substituting the given values to find the elastic constant to get:
Therefore, the elastic constant is 2 Newton/cm.
Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:
With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:
This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.