College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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A force acts on a particle, which is initially at rest; the force is F=(3i+4j)N. The particle is initially at di=(0.25i+0.50j)m. After t=2.0s, the particle is located at df=(3.25i+4.50j)m. 

a) How much work has been done by the force?

b) What is the power provided by this force?

c) What is the magnitude of the final velocity for the particle?

Expert Solution
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Step 1

(a)The displacement of the particle is given by,

Δd=df-di=3.25i^+4.50j^m-0.25i^+0.50j^m=3.00i^+4.00j^m

The work done by the force is,

W=F·Δd=3i^+4j^N·3.00i^+4.00j^m=9+16J=25 J

(b)

The power provided by the force is,

P=Wt

Here, t is the time for which the force acts.

Substituting,

P=25.00 J2.0 s=12.5 W

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