A block of mass √2kg is released from the top of an inclined smooth

1.	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?
Answer» 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.

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?

Answer»

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.

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