Calculus based physics problem. (energy and fricition)

[oohaysomeone]

In the figure, a 3.8 kg block slides along a track from one level to a higher level after passing through an intermediate valley (frictionless). The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 6.9 m/s, the height difference is h = 1.2 m, and μk = 0.649. Find d.

i've already found that energy transferred to thermal energy is 45.771 J, the normal force = 37.24 and the KE frictional force = 24.168 J but I'm not sure what to do with these numbers.

 

[oohaysomeone]

I think you use W = fd. But, I'm not sure how.

 

[baba89]

Based on the given information, we can use the conservation of energy principle to solve for the distance d. Initially, the block has kinetic energy (KE) equal to 1/2*m*v0^2, where m is the mass of the block and v0 is the initial speed. As the block moves up the incline, it gains potential energy (PE) equal to m*g*h, where g is the acceleration due to gravity and h is the height difference. At the top of the incline, the block has a total energy equal to KE + PE.

Now, as the block moves on the horizontal surface with friction, it loses some of its energy due to the frictional force. The work done by friction is equal to the frictional force multiplied by the distance d. Therefore, we can set up the equation:

1/2*m*v0^2 + m*g*h = μk*m*g*d

Solving for d, we get:

d = (1/2*v0^2 + g*h)/(μk*g)

Substituting the given values, we get:

d = (1/2*6.9^2 + 9.8*1.2)/(0.649*9.8) = 2.51 m

Therefore, the distance d that the block travels on the horizontal surface before coming to a stop is 2.51 meters.