help in spring question physics?
A block of mass 1.5 kg is placed on a flat surface, and it is being pulled horizontally
by a spring with a spring constant 1.2 × 103 N/m The coefficient
of static friction between the block and the table is µs = 0.60, and the coefficient of sliding
friction is µk = 0.40.
(a) By what amount must the spring be stretched to start the block moving?
(b) What is the acceleration of the block if the stretch of the spring is maintained at a constant value equal to that required to start the motion?
(c) By what amount must the spring be stretched to keep the mass moving at constant
- AshLv 77 months agoFavorite Answer
(a) To start the block moving, the static friction should be overcome
spring force = Static frictional force
kx = μ(s)N
kx = μ(s)mg
x = μ(s)mg/k
x = 0.60*1.5*9.8/(1.2 x 10³)
x = 0.00735 m or 7.35 mm
The block should be stretched just over 7.35 mm to start moving
(b) F(net) = Spring force - Kinetic frictional force
ma = kx - μ(k)N
ma = kx - μ(k)mg
a = kx/m - μ(k)g
a = (1.2 x 10³)*(0.00735)/1.5 - 0.40*9.8
a = 2.0 m/s²
(c) To keep constant speed, the net force on the block must be 0
F(net) = Spring force - Kinetic frictional force
0 = kx - μ(k)N
kx = μ(k)mg
x = μ(k)mg/k
x = 0.40*1.5*9.8/(1.2 x 10³)
x = 0.0049 m or 4.9 mm
The spring should be kept stretched by 4.9 mm to maintain constant speed.