Question

A 5 kg block is on a horizontal surface and is attached to a spring as shown. the spring has a spring constant of 700 n/m. at the instant shown, the spring is compressed by 2 cm, and the block is sliding toward the left. the coefficient of kinetic friction between the block and the surface is 0.12. what is the acceleration of the block at this instant?

Answer 1

The acceleration of the block at this instant is approximately 4.18 m/s² to the left.

Step-by-step explanation:

The acceleration of the block can be determined through the following steps. First, consider the forces acting on the block. The spring force, given by Hooke's Law (F_spring = -kx), opposes the direction of compression. The kinetic friction force (F_friction = μ_k N) acts opposite to the direction of motion.

The net force (F_net) is the sum of the spring force and the friction force. Applying Newton's second law (F_net = ma) and considering the negative sign for forces opposing motion, the equation becomes ma = -kx - μ_k x N. The normal force (N) is equal to the gravitational force acting on the block (N = mg).

Substituting this into the equation, we get ma = -kx - μ_k x mg. Solving for acceleration (a), we find a = (-kx - μ_k x mg) / m. Plugging in the given values, we get a = (-700 x 0.02 - 0.12 x 5 x 9.8) / 5, resulting in an acceleration of approximately 4.18 m/s² to the left.