A 800-g block is dropped onto a vertical spring with spring constant k =210.0 N/

Question

A 800-g block is dropped onto a vertical spring with spring constant k =210.0 N/m. The block becomes attached to the spring, and the spring compresses 57.1 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by its weight? While the spring is being compressed, what work is done on the block by the spring force? What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

work done by weight= m*g*h

= 0.8 Kg*9.8 m/s^2 * 0.571 m

= 4.48 J

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work done by spring force = - 0.5*k*x^2 {negative because force and displacement in different direction}

= - 0.5*210N/m * (0.571)^2

= - 34.23 J

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Kinetic energy at to top + work done by gravity + work done by spring = 0

0.5*m*v^2 +4.48 - 34.23 = 0

0.5*0.8*v^2 = 29.75

v= 8.62 m/s

Answer:8.62m/s

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if the speed is doubled,i.e speed becomes, 8.62*2 = 17.24 m/s

kinetic energy = 0.5*m*v^2 = 0.5*0.8*17.24^2 = 118.89 J

let compression be x

Kinetic energy at to top + work done by gravity + work done by spring = 0

118.89 + m*g*x - 0.5*k*x^2 = 0

118.89 + 0.8*9.8*x - 0.5*210*x^2 = 0

105 x^2 - 7.84 x - 118.89 = 0

After solving for x we get, x=1.1 m and x= -1.02 m but x can't be negative.

so, x= 1.02 m

so compression will be 1.02 m or 102 cm

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