Question

A 5 kg block is sliding on rough horizontal surface with speed 4 m/s at the moment when it just starts compressing a uncompressed spring of the spring constant 100 N/m . If the kinetic friction force between the block ans surface is 10 N, then the spring is maximum compressed by....? ANS== 4/5

Answer 1

$\bigstar$ $\sf{\green{\underline{\underline{\blue{To\:Find:-}}}}}$

• The Maximum Compression(x) of Spring .

$\bigstar$ $\sf{\green{\underline{\underline{\blue{SOLUTION:-}}}}}$

☞ GIVEN:-

• Mass of Block (m) = 5 kg
• Velocity of Block (v) = 4 m/s
• Spring Constant (k) = 100 N/m
• The kinetic friction force between the block and surface = 10 N

☞ See the above attachment picture .

☞ We have know that,

• Change in Kinetic energy = work done by all forces .

(╬) Here ,

$\tt{\red{\boxed{\underline{\:\:\:\:{\triangle{K\:=\:W_s\:+\:W_f\:\:\:\:}}}}}}$

☞ where ,

• ∆K = change in Kinetic energy = 1/2×mv²

• $\tt{\:W_s\:=\:Work\:done\:by\:spring\:\:=\:{\dfrac{1}{2}}kx^2\:}$

• $\tt{\:W_f\:=\:Work\:done\:by\:friction\:\:=\:(force\:\times\:displacement)\:}$

☞ Now,

$\tt{\:{\dfrac{1}{2}}mv^2\:=\:{\dfrac{1}{2}}kx^2\:+\:(force\times{displacement})\:}$

$\tt{\implies\:{\dfrac{1}{2}}\times{5}\times{4^2}\:=\:{\dfrac{1}{2}}\times{100}\times{x^2}\:+\:10\times{x}\:}$

$\tt{\implies\:50x^2\:+10x\:-\:40\:=\:0\:}$

➜ To solve the above quadratic equation we get,

$\tt{\implies\:x\:=\:{\dfrac{4}{5}}\:or\:{-1}\:}$

[Note:- Distance is always Positive]

$\tt{\purple{\boxed{\implies\:x\:=\:{\dfrac{4}{5}}\:m\:}}}$

☞ Therefore, The spring is maximum compresses by “4/5 m”.

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:({\dfrac{4}{5}})\:meter\:}}}}$

Answer 2

$\green{ \fcolorbox{grey}{grey}{ \checkmark \: \textsf{Verified \: answer}}}$

The spring is maximum compresses by “4/5 m”.