A 2.45 block on a horizontal floor is attached to a horizontal spring that is in

Question

A 2.45 block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0360 . The spring has force constant 800 . The coefficient of kinetic friction between the floor and the block is 0.44 . The block and spring are released from rest and the block slides along the floor.
What is the speed of the block when it has moved a distance of 0.0170 from its initial position? (At this point the spring is compressed 0.0190 .)

Explanation / Answer

Initial energy stored in the spring = 1/2*(k)*(x^2) =1/2*(800)*(0.036^2) = 0.5184 Now in the given situation, the spring is compressed by 0.019 So energy released = 1/2*(800)*(0.017^2) = 0.1156 Work done against friction = µ*N*x = µ*m*g*x = (0.44)*(2.45x10^-3)*(9.8)*(0.017) = 0.1796 x 10^-3 =>Kinetic energy of block = 1/2*m*(v^2) = 0.1156 - 0.1796 x 10^-3 = 0.1154204 =>v^2 = 94.2207 => v= 9.70

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