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Force F(x)=(10N)sin2πt4sis exerted on a 250 g particle during the interval0s≤t≤2.0s. If the particle starts from rest, what is its speed at t = 2.0 s?

Short Answer

Expert verified

The speed att=2.0secis50.92m/s

Step by step solution

01

Step 1. Given information

Force expression, F(x)=(10N)sin2Ï€t4s

The time variation of force value.

02

Step 2. Apply and calculate

We apply the impulse-momentum theorem,

∫0210Nsin2πt4=0.25×vfinal−42πcosπ−cos0×10=0.25×vfinalvfinal=50.92m/s

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