Question

A 16.4-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 4.50 m/s in 1.55 s. In the process, the spring is stretched by 0.236 m. The block is then pulled at a constant speed of 4.50 m/s, during which time the spring is stretched by only 0.0565 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Answer 1

mass of block = m = 16.4 kg

k = spring constant

f = frictional force

V_i = initial velocity = 0 m/s

V_f = final velocity = 4.50 m/s

t = time taken = 1.55 sec

a = acceleration = (V_f- V_i) /t = (4.50 - 0 ) / 1.55 = 2.9 m/s²

the spring force acting on the block is given as ::

F_s = kX

X = stretch in the spring.

when spring expansion is , X = 0.236 m, the force equation can be given as ::

F_s - f = ma                   (since the block moves at uniform acceleration)

kx - f = (16.4) (2.9)

K (0.236) - f = 47.56                     Eq-1

when spring is expansion , X = 0.0565 m, the force equation can be given as ::

F_s - f = m(0)                   (since the block moves at constant speed)

kx - f = 0

kx = f

f = k (0.0565)                          Eq-2

Using eq-1 and Eq-2

K (0.236) - k (0.0565) = 47.56  

K = 264.96 N/m

b) Using Eq-2

f = k (0.0565)   

f = (264.96) (0.0565)   

f = 14.97 N

frictional force on the block on flat surface is given as ::

u_k mg = f

inserting the values

u_k = f / mg = 14.97 / (16.4 x9.8 )

u_k = 0.093