1. Fnet=ma=Fs
Fnet=ma=usFn
Fnet=ma=usmg
mass canceles
a=usg
2. not a very good visual description, cant figure out what you mean. draw pictures. use a free body diagram
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?
2 answers
Q1:
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.Q2:
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.
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.Q2:
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.
1 answer