A block of mass root 2 kg is released

Asked by admin @ 23/08/2021 in Physics viewed by 329 People

A block of mass root 2 kg is released from the top of an inclined smooth surface as shown in fig. if spring constant of spring is 100n/m and block comes to rest after compressing the spring by 1m. then distance travelled befpre it comes to rest

Answered by admin @ 23/08/2021

_____________________
At the bottom of the incline:
_____________________

Energy lost by the block as the spring slows it to a stop (the block's kinetic energy at the bottom of the incline) = Energy gained by the spring

= (1/2)(100N/m)(1m)^2 = 50J

____________________________
As the block slides down the incline:
____________________________

Energy at the top = Energy at the bottom

Ki + Ui = Kf + Uf

Ui = Kf

mgh = 50 J where h is the height of the incline (the altitude)

(√2kg)(9.81m/s^2)(h) = 50J

h = 50/(√2x9.81) m

h = 50/13.87 m

h = 3.6 m

The incline is a right triangle with height h (the altitude) and hypotenuse d (the plane length).

Sin45o = h/d,

Then, d = 5m.

Hence, the distance travelled by block before it comes to rest is 5m.

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