Ablock of mass m = 1.00 kg is attached to a spring of force constant k = 500 n/m. the block is pulled to a position xi= 5.00 cm to the right of equilibrium and released from rest. find the speed the block has as it passes through equilibrium if the coefficient of friction between block and surface is 0.350.

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

Put the value into the formula

Hence, The speed of the block is 4.96 m/s.

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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