posted by ericka .
Question # 1 : What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.82?
Question # 2 : A 1.5-kg block rests on top of a 7.5kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere.
Part a) What force F must be applied to the bottom block so the top block accelerates to the right at 2.9 m/s^2 ?
Part b) What is the tension in the connecting cord?
The car moves forward by the reaction force from the ground, produced due to friction between tires and road. According to the Newton’s 3rd law
F12 = F21
F(net)max = F(friction)max
m•a = k• m•g
a = k• g = 0.82•9.8 = 8.04 m/s^2
m1 =1.5 kg, m2 = 7.5 kg, a=2.9 m/s²
(a) The horizontal projections of the equations of motion for each block are
m1•a = T,
m2•a = T-F,
F = (m1+m2) •a = (1.5+7.5) •2.9 = 26.1 N,
(b) T= m2•a - F= 7.5•2.9 – 26.1 = 4.35 N.