Question

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

Answer 1

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At the bottom of the incline: 
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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 

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As the block slides down the incline: 
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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.

Answer 2

Answer:

Explanation:

As we know by law of COME spring energy will be equal to total work done so by equating equations :1/2×200×1^2=f.x

Force is mg sin30(incline plane)and after solving we get x=2.5