Question

A 11.4-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 4.08 m/s in 1.13 s. In the process, the spring is stretched by 0.231 m. The block is then pulled at a constant speed of 4.08 m/s, during which time the spring is stretched by only 0.0543 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Answer 1

a)

consider the case when the block accelerate :

m = mass of the block = 11.4 kg

v_i = initial speed of block = 0 m/s

v_f = final speed of block = 4.08 m/s

t = time taken by the block = 1.13 sec

acceleration of the block is given as

a = (v_f - v_i )/t

a = (4.08 - 0)/1.13

a = 3.6 m/s²

k = spring constant

x = stretch in the spring = 0.231 m

F = tension force in the spring = k x = (0.231) k

f = kinetic frictional force acting on the block

force equation for the motion of the block is given as

F - f = ma

(0.231) k - f = (11.4) (3.6)

(0.231) k - f = 41.04 eq-1

consider the case when block moves at constant speed :

a' = acceleration of block = 0 m/s²

k = spring constant

x' = stretch in the spring = 0.0543 m

F' = tension force in the spring = k x' = (0.0543) k

f = kinetic frictional force acting on the block

force equation for the motion of the block is given as

F' - f = ma'

(0.0543) k - f = (11.4) (0)

(0.0543) k = f   eq-2

using eq-1 and eq-2

(0.231) k - (0.0543) k = 41.04

k = 232.3 N/m

b)

using eq-2

(0.0543) k = f

(0.0543) (232.3) = f

f = 12.6 N

u_k = Coefficient of kinetic friction

kinetic frictional force

f = u_k mg

12.6 = u_k (11.4 x 9.8)

u_k = 0.113