A 20.0 g object is placed against the free end of a spring (with spring constant

Question

A 20.0 g object is placed against the free end of a spring (with spring constant k equal to 25.0 N/m) that is compressed 10.0 cm. Once released, the object slides (with friction) across the tabletop and eventually lands 0.96 m from the edge of the table on the floor, as shown in the figure. The tabletop is 1.00 m above the floor level.

What is the speed of the block at the moment it leaves the tabletop? Assume air drag is negligible (m/s).

How much mechanical energy is lost from the block-spring system during the sliding motion? Give energy lost as a positive value.

Explanation / Answer

let t is the time taken to fall down from table top.

Apply, h = 0.5*g*t^2

t = sqrt(2*h/g)

= sqrt(2*1/9.8)

= 0.4517 s

so, speed of the block at the moment it leaves the tabletop, v = x/t

= 0.96/0.4517

= 2.125 m/s <<<<<----------Answer

Mechaical energy lost = 0.5*k*x^2 - 0.5*m*v^2

= 0.5*25*0.1^2 - 0.5*20*10^-3*2.125^2

= 0.0798 J <<<<<----------Answer

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