A 58.0-kg skier starts from rest at the top of a ski slope of height 70.0 .If frictional forces do −1.09×104 of work on her as she descends, how fast is she going at the bottom of the slope?

I will have to assume the height is in meters and the friction work is in Joules. Don't be sloppy by omitting units.

Her kinetic energy will be
(M*V^2)/2 = M g H - 1.09*10^4 J

You know M, g and H.

Solve for V

To find the speed of the skier at the bottom of the slope, we can apply the principle of conservation of energy. The initial potential energy of the skier at the top of the slope is equal to the sum of the final kinetic energy and the work done by friction.

1. Calculate the potential energy at the top of the slope using the formula: PE = mgh, where m is the mass of the skier (58.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the slope (70.0 m).
PE = 58.0 kg * 9.8 m/s^2 * 70.0 m = 39,956 J

2. Calculate the work done by friction (W) using the given value: W = -1.09 × 10^4 J.

3. The final kinetic energy (KE) of the skier at the bottom of the slope can be found by subtracting the work done by friction from the initial potential energy: KE = PE - W.
KE = 39,956 J - (-1.09 × 10^4 J) = 50,056 J

4. Calculate the final velocity (v) using the formula for kinetic energy: KE = 1/2 mv^2, where m is the mass of the skier and v is the final velocity.
50,056 J = 1/2 * 58.0 kg * v^2

5. Rearrange the equation and solve for v:
v^2 = (2 * 50,056 J) / 58.0 kg
v^2 ≈ 172,591 m^2/s^2
v ≈ √172,591 m/s
v ≈ 415.3 m/s

Therefore, the skier is going approximately 415.3 m/s at the bottom of the slope.

To find the speed of the skier at the bottom of the slope, we can use the principle of conservation of energy.

The initial potential energy (U) at the top of the slope is given by:

U = m * g * h

Where:
m = mass of the skier = 58.0 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the slope = 70.0 m

U = 58.0 kg * 9.8 m/s^2 * 70.0 m
U = 39892 J (Joules)

The work done by friction (W) is given as -1.09 × 10^4 J (negative because work is done against the motion).

Using conservation of energy, the total mechanical energy is conserved.

Initial mechanical energy (Ei) = Final mechanical energy (Ef)

Ei = U (initial potential energy)
Ef = 1/2 * m * v^2 (final kinetic energy)

Ei = Ef + W

Solving for v (speed of the skier at the bottom):

1/2 * m * v^2 = U + W

v^2 = (U + W) * 2 / m

v = sqrt((U + W) * 2 / m)

Plugging in the values:

v = sqrt((39892 J - 1.09 × 10^4 J) * 2 / 58.0 kg)
v = sqrt((39892 J - 10900 J) * 2 / 58.0 kg)
v = sqrt(28992 J * 2 / 58.0 kg)
v = sqrt(996 J / kg)
v = sqrt(996) m/s
v ≈ 31.6 m/s

Therefore, the skier is going at approximately 31.6 m/s at the bottom of the slope.