4. The block shown is released from rest when the spring is stretched a distance

Question

4. The block shown is released from rest when the spring is stretched a distance d. If k = 50 N/m, m = 0.50 kg, d = 10 cm, and the coefficient of kinetic friction between the block and the horizontal surface is equal to 0.25, determine the speed of the block when it first passes through the position for which the spring is unstretched Using the Work energy theorem work is done by the spring and the friction force kxf2- kx2 + ?.mgd cos 180- mv.2-? my' 0 + 2 (50) (0.)(0.25) (0.5) (9.8) (0.1) (-1) (0.5) (v) 0 0 (50) (0.01)+ (0.25) (0.5) (9.8) (0.1) -1) 2 (0.5) (v)+0 0.25 (-0.1225) 0.25 (v') 0.1275 0.25 (v') 0.1275 0.25 (v2) a. 92 cm/s b. 61 cm/s c. 71 cm/s d. 82 cm/s e. 53 cm/s Convert cm to m f* d cos ? is the formula for friction or why do you use this formula? 10 cm xin 0.1 m 100m When the block is at rest the initial kinetic energy is zero. The spring will have potential energy. When the block is released the potential energy will transfer to kinetic energy (final kinetic energy).The spring will have zero final potential energy 0.25 0.25 .51 = v2 0.71mv Convert to cm 71 0.71 m x100 cm71 cm 1 m

Explanation / Answer

Yes, above used method is Absolutely correct.

final answer will be v = 0.71 m/sec = 71 cm/sec

Now about your question in Red

We know that work done is given by:

W = F.d = F*d*cos theta

F = applied force

d = distance traveled

theta = angle between direction of force and displacement

Now we need to calculate Work done by friction force, So

Wf = f*d*cos theta

f = friction force = uk*N = uk*m*g

N = normal force on object

theta = 180 deg = because friction is always in opposite direction of motion

So

Wf = uk*m*g*d*cos 180 deg.

Let me know if you still have any doubt.

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