A block of mass √2 kg is released from the top
of an inclined smooth surface as shown in figure.
If spring constant of spring is 100 N/m and block
comes to rest after compressing the spring by
1 m, then the distance travelled by block before it
comes to rest is
Answers
Answered by
6
Explanation:
energy lost by the block = energy gained by the spring.
energy gained =1/2 x 100 x 1^2= 50J
energy of the block was mgh
mgh=50J
h=50/10*√2
=5/√2m
now, sin45°=perpendicular/hypotenuse
hypotenuse=(5/√2)/(1/√2)
=5m
therefore ,before coming to rest .
the block travels 5m + 1m(while compressing the spring).
total distance travelled= 6m
Answered by
0
Answer:
energy lost by the block = energy gained by the spring.
energy gained =1/2 x 100 x 1^2= 50J
energy of the block was mgh
mgh=50J
h=50/10*√2
=5/√2m
now, sin45°=perpendicular/hypotenuse
hypotenuse=(5/√2)/(1/√2)
=5m
therefore ,before coming to rest .
the block travels 5m + 1m(while compressing the spring).
total distance travelled= 6m
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