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Chapter 7: Problem 87

Within cells, small organelles containing newly synthesized proteins are transported along microtubules by tiny molecular motors called kinesins. What force does a kinesin molecule need to deliver in order to accelerate an organelle with mass \(0.01 \mathrm{pg}\left(10^{-17} \mathrm{kg}\right)\) from 0 to \(1 \mu \mathrm{m} / \mathrm{s}\) within a time of \(10 \mu \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The force exerted by a kinesin molecule on the organelle is \(10^{-18}\,\mathrm{N}\).

Step by step solution

01

Identify the given values

The given values are: - Mass of the organelle (m): \(0.01\,\mathrm{pg} = 10^{-17}\,\mathrm{kg}\) - Initial velocity (v0): \(0\,\frac{\mu\mathrm{m}}{\mathrm{s}}\) - Final velocity (vf): \(1\,\frac{\mu\mathrm{m}}{\mathrm{s}} = 10^{-6}\,\frac{\mathrm{m}}{\mathrm{s}}\) - Time interval (t): \(10\,\mu\mathrm{s} = 10^{-5}\,\mathrm{s}\)
02

Calculate the acceleration

To find the acceleration (a), we can use the formula a = (vf - v0) / t. a = \(\frac{10^{-6}\,\frac{\mathrm{m}}{\mathrm{s}} - 0\,\frac{\mu\mathrm{m}}{\mathrm{s}}}{10^{-5}\,\mathrm{s}}\) a = \(10^{-1}\,\frac{\mathrm{m}}{\mathrm{s}^2}\)
03

Calculate the force

Now that we have the acceleration, we can use Newton's second law (F = m * a) to find the required force (F). F = \((10^{-17}\,\mathrm{kg})(10^{-1}\,\frac{\mathrm{m}}{\mathrm{s}^2})\) F = \(10^{-18}\,\mathrm{N}\) So, a kinesin molecule needs to deliver a force of \(10^{-18}\,\mathrm{N}\) in order to accelerate the organelle as described in the problem.

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Most popular questions from this chapter

A projectile of 1.0-kg mass approaches a stationary body of \(5.0 \mathrm{kg}\) at \(10.0 \mathrm{m} / \mathrm{s}\) and, after colliding, rebounds in the reverse direction along the same line with a speed of $5.0 \mathrm{m} / \mathrm{s} .$ What is the speed of the 5.0 -kg body after the collision?
Verify that the SI unit of impulse is the same as the SI unit of momentum.
A jet plane is flying at 130 m/s relative to the ground. There is no wind. The engines take in \(81 \mathrm{kg}\) of air per second. Hot gas (burned fuel and air) is expelled from the engines at high speed. The engines provide a forward force on the plane of magnitude \(6.0 \times 10^{4} \mathrm{N} .\) At what speed relative to the ground is the gas being expelled? [Hint: Look at the momentum change of the air taken in by the engines during a time interval \(\Delta t .\) ] This calculation is approximate since we are ignoring the \(3.0 \mathrm{kg}\) of fuel consumed and expelled with the air each second.
A 0.020-kg bullet is shot horizontally and collides with a \(2.00-\mathrm{kg}\) block of wood. The bullet embeds in the block and the block slides along a horizontal surface for \(1.50 \mathrm{m} .\) If the coefficient of kinetic friction between the block and surface is \(0.400,\) what was the original speed of the bullet?
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