An 85-kg skier slides down a frictionless ramp of snow at an initial height of 15 meters above ground level and a speed of 10 m/s, as shown in the diagram below: ↑ Unpacked Snoww When the skier reaches the ground, they slide through a horizontal plane of 25 meters of unpacked snow that has some friction. If the skier travels at 15 m/s after sliding past the patch of unpacked snow, what is the magnitude of the unpacked snow's friction in newtons? Round to the nearest integer.

While skiing down the ramp, gravitational potential energy is converted into kinetic energy.
While skiing through the unpacked snow on the horizontal plane, some energy is lost because of the work that friction did on the skier.

The mechanical energy of an object is the sum of its potential energy and kinetic energy. Under the assumption that the ramp is frictionless, mechanical energy would be conserved while skiing down the ramp. Friction would be responsible for any of the difference between the initial mechanical energy and the mechanical energy after traveling through the horizontal plane.

Hence, approach this question in the following steps:

Find the difference between the sum of kinetic energy and gravitational potential energy when on the top of the ramp, versus after the travelling through the unpacked snow. This energy difference is equal in magnitude to the work that friction did on the skier.
Divide the work done by the distance over which the friction force was applied to find the magnitude of friction.

If the ground level is at height [tex]0\; {\rm m}[/tex], the top of the ramp would be at a height of [tex]15\; {\rm m}[/tex]. Mechanical energy on the top of the ramp consists of:

Gravitational potential energy:
[tex]m\, g\, h \approx (85\; {\rm kg})\, (9.81\; {\rm m\cdot s^{-2})\, (15\; {\rm m}) \approx 12\, 507.75\; {\rm J}[/tex].
Kinetic energy:
[tex]\displaystyle \frac{1}{2}\, m\, v^{2} = \frac{1}{2}(85\; {\rm kg})\, (10\; {\rm m\cdot s^{-1}})^{2} \approx 4\, 250\; {\rm J}[/tex].

Thus, the mechanical energy of the skier while at the top of the ramp would be approximately:

[tex]12\, 507.75\; {\rm J} + 4\, 250\; {\rm J} \approx 16\, 757.75\; {\rm J}[/tex].

Similarly, the mechanical energy after travelling through the plane at ground level would consist of:

Gravitational potential energy:
[tex]m\, g\, h \approx (85\; {\rm kg})\, (9.81\; {\rm m\cdot s^{-2})\, (0\; {\rm m}) = 0\; {\rm J}[/tex].
Kinetic energy:
[tex]\displaystyle \frac{1}{2}\, m\, v^{2} = \frac{1}{2}(85\; {\rm kg})\, (15\; {\rm m\cdot s^{-1}})^{2} \approx 9\, 562.5\; {\rm J}[/tex].

The mechanical energy of the skier at that moment would be approximately:

[tex]0\; {\rm J} + 9\, 562.5\; {\rm J} = 9\, 562.5\; {\rm J}[/tex].

The difference between the mechanical energy at the top of the ramp and at ground level would be approximately:

[tex]16\, 757.75\; {\rm J} - 9 \,562.6\; {\rm J} \approx 7\,195.15\; {\rm J}[/tex].

This energy difference would be the same as the magnitude of work that friction exerted on this skier across the distance of [tex]25\; {\rm m}[/tex]. To find the magnitude of friction on the skier, divide the magnitude of work by distance across which this force is exerted:

[tex]\displaystyle (\text{force}) = \frac{(\text{work})}{(\text{distance})} \approx \frac{7\, 195.15\; {\rm J}}{25\; {\rm m}} \approx 287.81\; {\rm N}[/tex].