A 10lb pendulum is released from rest and strikes a 50lb blo

A 10-lb pendulum is released from rest and strikes a 50-lb block with e = 0.7. The block slides on a frictionless surface. Find: (0.25) The velocity of the pendulum at impact, in ft/s. (0.25) The tension of the cord at impact, in lb. (0.25) The block\'s velocity immediately after impact, in ft/s. (0.25) The spring constant to stop the block with 6 inches of deflection, in lb/ft.

Solution

a)

T₁+V₁ =T₂+V₂

0+10*3 =(1/2)(10/32.2)*V_p²+0

V_p=13.9 ft/s

b)

T-W =ma_n

=>T=10+(10/32.2)(13.9²/3)

T=30 lb

c)

(1/2)m_pV_p1+(1/2)m_BV_B1=(1/2)m_pV_p2+(1/2)m_BV_B2

0+(10/32.2)(13.9)=(10/32.2)V_p2+(50/32.2)V_B1

10V_p2+50V_B2=139------------------------------------------1

(V_B2-V_p2) =e(V_p1-V_B1) =0.7(13.9-0)=9.73

V_p2=V_B2-9.73

From 1

10(V_B2-9.73)+50V_B2=139

V_B2 = 236.3/60 =3.94 ft/s

d)

1 inch =0.08333 ft

T₁+V₁ =T₂+V₂

(1/2)(50/32.2)*3.94²+0 =0+(1/2)*K*(6*0.08333)

K=96.4 Lb/ft