A block of mass m = 4 kg slides along a horizontal table when it encounters the free end of a horizontal spring of spring constant k = 14 N/m. The spring is initially on its equilibrium state, defined when its free end is at x=0 . Right before the collision, the block is moving with a speed vi = 5 m/s . There is friction between the block and the surface. The coefficient of friction is given by μ = 0.86 . How far did the spring compress when the block first momentarily comes to rest?

KE = (1/2)(m)(v)^2

= (0.5)(4)(5)^2
= 50 J

Constant friction force encountered:

N = (0.94)(4)(10)
= 37.6 N

Energy absorbed by friction:

Fd = 37.6x

Energy absorbed by the spring:

=(1/2)(17)(x)^2
=8.5x^2

Final Equation:

50 = 8.5x^2+37.6x
0 = 8.5x^2+37.6x-50

Solve the quadratic you will get the answer!!

i am not getting the correct answer

your values are wrong, but i correct it

To find the distance the spring compresses when the block first momentarily comes to rest, we need to use the principles of mechanical energy conservation.

1. First, calculate the gravitational potential energy (Ug) of the block at its initial position on the table. The gravitational potential energy is given by Ug = mgh, where m is the mass of the block (4 kg) and h is the height (which is assumed to be zero in this case).

Ug = mgh = 4 kg * 9.8 m/s^2 * 0 m = 0 J

2. Next, calculate the initial kinetic energy (Ki) of the block just before the collision. Kinetic energy is given by Ki = 0.5mv^2, where m is the mass of the block (4 kg) and v is the initial velocity (5 m/s).

Ki = 0.5 * 4 kg * (5 m/s)^2 = 0.5 * 4 kg * 25 m^2/s^2 = 50 J

3. Since the block comes to rest momentarily, all of its kinetic energy is transferred into potential energy stored in the spring (Us). Therefore, the initial potential energy stored in the spring when the block comes to rest is equal to the initial kinetic energy of the block.

Us = Ki = 50 J

4. The potential energy stored in a spring is given by Us = 0.5kx^2, where k is the spring constant (14 N/m) and x is the displacement of the spring from its equilibrium position.

0.5kx^2 = 50 J

5. Rearranging the equation, we have x^2 = (50 J) / (0.5k), and solving for x gives us:

x = √[(50 J) / (0.5k)]

Substituting the given values, we get:

x = √[(50 J) / (0.5 * 14 N/m)] = √[(50 J) / (7 N/m)] = √(7.14) m ≈ 2.67 m

Therefore, the spring compresses approximately 2.67 meters when the block first momentarily comes to rest.