Question

(5\%) Problem 14: A spring with a spring constant of k=191 N/m is initially compressed by a block a distance d=0.23 m from its unstretched length. The block is on a horizontal surface with coefficients of static and kinetic friction μ1​ and μk​ and has a mass of m=2 kg. Refer to the figure. Assuming the block has just begun to move and the coefficient of kinetic friction is μk​=0.2, what is the block's acceleration in meters per second squa a= Hints: If for a 0. deduction. Hiats remainisg 0. Pecdbackt

Answer 1

The acceleration of the block is approximately 20 m/s^2.

The acceleration of the block can be determined using Newton's second law of motion, as you've correctly outlined.

a = (kx - μmg) / m

where;

• m is the mass of the block

• k is the spring constant

• x is the compression of the spring

• g is acceleration due to gravity

• μ is the coefficient of friction

a = ((191 N/m × 0.23 m) - (0.2 × 2 kg × 9.8 m/s2)) / 2 kg

a = (43.93 - 3.92) / 2

a = 40.01 / 2

a = 20.005 m/s2

Therefore, the correct acceleration of the block is approximately 20 m/s2.

Answer 2

The acceleration of the block is determined as 20 m/s².

How to calculate the acceleration of the block?

The acceleration of the block is calculated by applying Newton's second law of motion;

F(net) = ma

F(elastic force) - F(friction) = ma

kx - μmg = ma

a = (kx - μmg)/m

where;

• m is the mass of the block
• k is the spring constant
• x is the compression of the spring
• g is acceleration due to gravity
• μ is the coefficient of friction

a = (kx - μmg)/m

a = (191 x 0.23 - 0.2 x 2 x 9.8 ) / 2

a = 20 m/s²