a block of mas \( m \) = 4.8 kg slides head on into a spring of spring constant \( k \) = 430 N/m. When the block stops, it has compressed the spring by 5.8 cm. The coefficient of kinetic friction between block and floor is 0.28. \( (g =9.8m/s^2) \)

"While the block is in contact with the spring and being brought to rest, what are (a)the work done by the spring force and (b) the increase in thermal energy of the blockfloor system? (c) What is the blocks speed just as it reaches the spring?"

According to this we need to calculate three values: Work, Thermal Energy and Speed of the block when it reaches the spring.

Let's do this by parts.

a) Work done by the spring:

In this case, we need to apply the following expression:

W = -1/2 kx² (1)

We know that k = 430 N/m, and x is the distance of compressed spring which is 5.8 cm (or 0.058 m). Replacing that into the expression:

W = -1/2 * 430 * (0.058)²

W = -0.7233 J

b) Increase in thermal energy

In this case we need to use the following expression:

ΔEt = Fk * x (2)

And Fk is the force of the kinetic energy which is:

Fk = μk * N (3)

Where μk is the coeffient of kinetic friction

N is the normal force which is the same as the weight, so:

N = mg (4)

Let's calculate first the Normal force (4), then Fk (3) and finally the chance in the thermal energy (2):

N = 4.8 * 9.8 = 47.04 N

Fk = 0.28 * 47.04 = 13.1712 N

Finally the Thermal energy:

ΔEt = 13.1712 * 0.058

ΔEt = 0.7639 J

c) Block's speed reaching the spring

As the block is just reaching the speed, the initial Work is 0. And the following expression will help us to get the speed:

V = √2Ki/m (5)

And Ki, which is the initial kinetic energy can be calculated with:

Ki = ΔU + ΔEt (6)

And ΔU is the same value of work calculated in part (a) but instead of being negative, it will be positive here. So replacing the data first in (6) and then in (5), we can calculate the speed:

Ki = 0.7233 + 0.7639 = 1.4872 J

Finally the speed:

V = √(2 * 1.4872) / 4.8

V = 0.7872 m/s

Hope this helps