Calculating Work and Energy in a Skier Problem

[triplel777]

Homework Statement

A 65-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.4 m/s. After coasting 1.9 m up the slope, the skier has a speed of is 3.2 m/s.

Homework Equations

The Attempt at a Solution

Work done if force F applied over a distance d.
W= Fd
Also
Ke= 0.5mV^2
Difference in Kinetic energy dKe
dKe=Ke1 - Ke2
dKe= 0.5m(V1^2 - V2^2)
dKe= W

a) W=dKe= 0.5m(V1^2 - V2^2)
W= 0.5x65(6.4^2 - 3.2^2)
W= 998.4 Joules

P=mgh
Pe= 65 x 9.81 x 1.9 sin(25)= 511.5 J

Wf= W- Pe= 998.4 - 511.5 = 486.9 J

what am i doing wrong?

 

[tiny-tim]

Hi triplel777!

A 65-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.4 m/s. After coasting 1.9 m up the slope, the skier has a speed of is 3.2 m/s.

I'm confused …

what's the question? ​

 

[triplel777]

(a) Find the work done by the kinetic frictional force that acts on the skis.

(b) What is the magnitude of the kinetic frictional force?

 

[tiny-tim]

Hi triplel77!

(try using the X² tag just above the Reply box; and we usually write KE and PE with capitals, and joules without )

A 65-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.4 m/s. After coasting 1.9 m up the slope, the skier has a speed of is 3.2 m/s.
…
a) W=dKe= 0.5m(V1^2 - V2^2)
W= 0.5x65(6.4^2 - 3.2^2)
W= 998.4 Joules

P=mgh
Pe= 65 x 9.81 x 1.9 sin(25)= 511.5 J

Wf= W- Pe= 998.4 - 511.5 = 486.9 J

what am i doing wrong?

Apart from making PE = 512.0 J, I get the same work done by friction as you do.

 

[triplel777]

huh... ok thanks

 

[triplel777]

my teacher said to use this formula
Wnc= delta KE+ delta PE
would this make any difference?

 

[gamer_x_]

delta PE is just your change in gravitational potential energy, delta KE is your change in kinetic energy. Does this formula seem different than the one you used?