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Sphinxa [80]
4 years ago
8

A very light ideal spring having a spring constant (force constant) of 8.2 N/cm is used to lift a 2.2-kg tool with an upward acc

eleration of 3.25 m/s2. If the spring has negligible length when it us not stretched, how long is it while it is pulling the tool upward?
Physics
1 answer:
vampirchik [111]4 years ago
6 0

Answer:

x = 3.5 cm

Explanation:

When spring is used as a lift tool

so the mass suspended on it will have spring force on it

So as per the force equation of mass we can say

F_{net} = ma

now net force on the mass is

F_{net} = kx - mg

F_{net} = kx - mg = ma

here we have

kx = mg + ma

now we have

x = \frac{mg + ma}{k}

x = \frac{2.2(9.8) + 2.2(3.25)}{8.2}

x = 3.5 cm

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A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s
expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

\vec v_f=\vec v_o+\vec a.t

\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is 2.7 m/s^2 to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

v_f=v_o+a.t

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\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s

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x=-7,41\ m

The motorboat ends up 7.41 meters to the west of the initial position

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