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Chapter 11: Q. 61 (page 290)

A 500 g particle has velocity vx=-5.0m/sat t=-2s. Force Fx=4-t2, where t is in s, is exerted on the particle betweent=-2sandt=2s. This force increases from 0 N at t=-2sto4Natt=0s and then back to 0Natt=2s. What is the particle’s velocity at t = 2 s?

Short Answer

Expert verified

The particle's velocity att=2secis16.33m/s

Step by step solution

01

Given information

We have given that a 500gparticle has velocityvx=-5.0m/satt=-2sand forceFx=4-t2N.

02

Analyze and apply

Applying the impulse-momentum theorem,

J=pfinal−pinitial∫−22Fxdt=mparitclevfinal−mparitclevinitial∫−224−t2dt=0.5kg×vfinal−0.5kg×−5m/s

On solving further,

4⋅2−−2−4323−(−2)3=0.5kg×vfinal+2.516−2.5=0.5kg×vfinalvfinal=16.33m/s

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