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A 0.380 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 28.0 N/m force constant that is at its equilibrium length. A 0.0600 kg wad of Play-Doh is thrown horizontally at the block with a speed of 2.60 m/s and sticks to it. Determine the amount in centimeters by which the Play-Doh-block system compresses the spring.

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Initial momentum of the Play-doh: 0.0600 x 2.60 = 0.156 kg/m/s

Total mass of the block and play-doh: 0.38 + 0.0600 = 0.44 kg.

Final momentum is mass x velocity = 0.44v

V = Initial momentum / mass

V = 0.156 / 0.44 = 0.3545 m/s

Work done by spring is equal to the Kinetic enrgy.

Work Done by spring = 1/2 *28.0 * distance^2 = 14 * d^2

KE = 1/2 * 0.44* 0.3545^2

set to equal each other:

14 * d^2 = 0.22 *0.12567

Solve for d:

d = √(0.22*0.12567)/14

d = 0.44 meters = 4.4cm

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