Incline, Friction, and Work: Calculating the Speed of a Snowboarder on a Hill

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A snowboarder descends a 50° incline with a coefficient of kinetic friction of 0.100, starting from rest and traveling 390 meters. The work done by friction is calculated to be -2.11 x 10^4 J. The final speed of the snowboarder is approximately 73 m/s or 264 km/h, although one response indicates this may be off by 3 m/s. Forum participants emphasize the importance of guiding users to understand the method rather than simply providing answers. The discussion highlights the balance between offering assistance and ensuring comprehension of the underlying physics concepts.
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A snowboarder descends a hill with an incline of 50°. the coefficient of kinetic friction of the board on the snow is µ = 0.100. The mass of the snowboarder and board is 86 kg. The snowboarder starts at rest and ends up 390 meters down the slope. (a) What is the work done on the snowboarder by the force of friction (watch sign)? (b) What is the boarder's speed at the end of the course?


Can anyone help me with this problem. I appreciate it.
 
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Welcome to the Forums,

One is expected to show one's efforts before asking for help.

HINT: Work - energy theorem
 
a) -2.11 x 10^4 J
b) 73m/s or 264km/h
 
fatra2 said:
a) -2.11 x 10^4 J
b) 73m/s or 264km/h
Please do not post answers, although they are correct (you second answer is off by 3 m/s); the OP learns nothing of the method. It is better to allow the poster to attempt their question then correct their method by guiding them to the correct answer.
 
Last edited:
Hi there,

I fully agree that giving out the complete answer is certainly not helping. I thought that having the final answer can be quite helpful in solving the problem. You can double check that you have the correct solution to the problem.
 
fatra2 said:
Hi there,

I fully agree that giving out the complete answer is certainly not helping. I thought that having the final answer can be quite helpful in solving the problem. You can double check that you have the correct solution to the problem.
Although I do agree with you, it can be useful to have an answer to work towards; it is the policy of this forum not give out answers since some 'unscrupulous' students may just copy down the answers without every really understanding the subject matter.

Your help is appreciated though, as you can see there are a lot of questions posted in the homework forums and anyone willing to help is much appreciated and I hope you will continue to do so.
 
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