A 1.50 kg mass slides to the right on a surface having a coefficient of kinetic

Question

A 1.50 kg mass slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.60). The object has a speed of v_i = 2.70 m/s when it makes contact with a light spring g. that has a force constant of 50.0 N/m. he object comes to rest after the spring has been compressed a distance d. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring. Find the distance of compression d. Find the speed v at the unstretched position when the object is moving to the left. m/s Find the distance D where the object comes to rest. m

Explanation / Answer

(a) initial kinetic energy = 1/2(mvi2 )

When compressed its potential energy= 1/2(kd2 )

Work done by friction = 0.25*mg*d

Thus

1/2(mvi2) =1/2(kd2 )+0.25*1.5*10*d

Here we took g=10m/sec2 ( for simplicity)

Putting the values of m and d and k we get

1/2(1.5*2.72)= 1/2(50*d2 )+0.25*15*d

Solving this we get quadratic equation in dd

5.5=25d2 +3.75ddSolving we get d= 0.5m(approx)

(b) when in equilibrium it has only kinetic energy

Same as initial kinetic energy -2wd

=5.5-2*1.5*10*0.5=9.5

So 1/2(mv2)= 9.5

So v=3.6 m/sec

(c) 3.6 = w*D= 1.5*10*D

D =3.6/15)=0.24m this is the distance when box comes to rest.

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