A block of mass √2kg is released from the top of an inclined smooth surface.If the spring constant is 100N/m

Mam as there is the limit of only 150 characters so I couldn't complete the question

A block of mass √2kg is released from the top of an inclined smooth surface.If the spring constant is 100N/m and block comes to rest after compressing the spring by 1m, then the distance travelled by block before it comes to rest is? This is the question Mam plz solve this it is very urgent

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Ok  Mam but Mam its answer is 5m and it is given that angle =45°

ok put the angle to 45 degree you will get the correct answer

I have replaced the value to angle 45 degree. You can make the answer as best

Mam but your h=3.6 so how come it become 5m? Mam maybe in calculating h there maybe some mistake Mam plz check plz mam

Ok Mam got it Thank u so much Mam

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/s²)(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). 

sin45^o = h/d, 

then d = 5m.

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