Problem 7.30 A 0.26-kg block is held in place against the spring by a 28-N horiz

Question

Problem 7.30 A 0.26-kg block is held in place against the spring by a 28-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v 1 = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v 2= 1.6 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 02. The velocity of the block is v 3 -1.4 m/s at C. The block moves on to D, where it stops. The initial compression of the spring is closest to: Part A V1 VA V2 smooth BoughE

Explanation / Answer

Data needed for the initial compression.

F = k*x = 28 N

v1 = 1.2 m/s

m = 0.26 kg

Now we know

½*m*v12 = ½*k*x2

x = (mv12)/kx

put the value of m, v1 and kx

x = (0.26*1.22)/28

x = 0.0133714 m

x = 1.3 cm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---