A 45kg sled is coasting with constant velocity at 3.0m/s over a patch of smooth and level ice. It enters a very rough section of ice that is 11m long in which the force of friction is 15N. With what speed will the sled emerge from the rough ice?

vf^2=Vi^2 + 2ad where a= force/mass, and it is negative here (it opposes motion). Solve for vf

a = F/m = 15N/45kg = 0.33 m/s^2 ,

vf^2 = vi^2 + 2ad
vf^2 = 3^2 + 2(0.33)(11)
vf^2 = 16.3
vf = 4m/s
is that correct?

To find the speed at which the sled will emerge from the rough ice, we can use the principle of conservation of mechanical energy. Since the sled is coasting with constant velocity, we know that the initial mechanical energy of the sled is equal to its final mechanical energy.

The initial mechanical energy (E1) of the sled can be calculated using the kinetic energy formula:

E1 = (1/2) * m * v1^2

where m is the mass of the sled (45 kg) and v1 is the initial velocity of the sled (3.0 m/s).

The final mechanical energy (E2) of the sled can be calculated using the work-energy principle:

E2 = E1 + W

where W is the work done by the force of friction on the sled.

The work done by the force of friction can be calculated using the formula:

W = F * d * cos(theta)

where F is the force of friction (15 N), d is the distance over which the force is applied (11 m), and theta is the angle between the force of friction and the displacement (0 degrees since the force of friction and displacement are in the same direction).

Substituting the given values into the formulas:

E1 = (1/2) * 45 kg * (3.0 m/s)^2 = 202.5 J

W = 15 N * 11 m * cos(0 degrees) = 165 J

Therefore, the final mechanical energy (E2) of the sled is:

E2 = 202.5 J + 165 J = 367.5 J

To find the final velocity (v2) of the sled, we can rearrange the kinetic energy formula:

E2 = (1/2) * m * v2^2

367.5 J = (1/2) * 45 kg * v2^2

Simplifying the equation:

v2^2 = (2 * 367.5 J) / 45 kg

v2^2 = 16.333...

Taking the square root of both sides:

v2 ≈ 4.04 m/s

Therefore, the sled will emerge from the rough ice with a speed of approximately 4.04 m/s.

a=F/m=15N/45kg=0.33 m/s^2

which in this case is Negative
vf^2=vi^2+2ad
vf^2=9+2(-.33)(11)
vf^2=9+(-7.26)
vf^2=1.74
vf=sqrt(1.74)
vf= 1.32 m/s