Question

A 2.65 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has force constant 830 N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . 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 m from its initial position? (At this point the spring is compressed 0.0130 m .)

Answer 1

mg = 0.03m. Mass of the block, m=2-65kg, spring is compressed for ge = 0.03 m, fu ilk spring constant, k=830 N/m. So, Enesgy stosed in the spring , E = 1 kg2.J Es įx830% (0.03)? = 0.3735 5 Mornal force on the block, M = mg. = 216.5*9.8 No.

So. N=25-97 N. Co-efficient of Rsiction, Mx = 0.45 so, Frictional force against the motion, Fu = UK N. 0:49 x 25-97 Ni . Fu = 11.6865 N Now, the block moves for a distance x=0.01fm from initial position, so, loss of energy due to fisichion, с. 5 хут - 11.686 5 x 0.017 = 0.1987 J. Now the energy of the block ire. the kinetic energy is E-E'), 'If u be the relocity So, then Imut= I mis E-E 2(E-E) m ve 2(E-5') -12260-3735 – 0.1987) 2x (0.3735 -0.1987) 2.65 U= 0.36 mes.