In our first example we will consider a very simple application of Newton’s seco

Q: In our first example we will consider a very simple application of Newton’s seco

The newton second law F = m a acceleration a = F/ m = 20 N /40kg = 0.5 m/s^2 b) acceleration a = F / m = 30 N / 40 kg = 0.75m/s^2 initial speed Vi = 0 , distance d = 0.34m V^2 - Vi^2 = 2 a d V^2 - 0 = 2 (0.75)(0.34) V = 0.714m/s

ID: 1331688 • Letter: I

Question

In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls with a constant horizontal force of magnitude 20 N on a box with mass 40 kg resting on the flat, frictionless surface of a frozen lake. What is the acceleration of the box?

a) If the box starts from rest and the worker pulls with a force of 30 N , what is the speed of the box after it has been pulled a distance of 0.34 m ?

Express your answer in meters per second to two significant figures.

Explanation / Answer

The newton second law

F = m a

acceleration

a = F/ m = 20 N /40kg = 0.5 m/s^2

acceleration a = F / m = 30 N / 40 kg = 0.75m/s^2

initial speed Vi = 0 , distance d = 0.34m

V^2 - Vi^2 = 2 a d

V^2 - 0 = 2 (0.75)(0.34)

V = 0.714m/s

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In our first example we will consider a very simple application of Newton\'s sec

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In our first example we will consider a very simple application of Newton’s seco

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