anonymous
  • anonymous
A boy coasting on a sled has a speed of 15 ft/s when he encounters a 30 degrees hill. The boy sled have a total weight of 60 lb and the coefficient of kinetic friction between sled and hill is .1. What is the boy's acceleration up the hill and how far does he go?
Physics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
can u solve this for me?
JamesJ
  • JamesJ
First, draw a diagram. When he is on the slope he has two forces acting on him: gravity \( F_g \) and friction \( F_f \). Create a coordinate frame where the x-axis is parallel to the hill and y is perpendicular to it. The normal force of the hill on the sled is in the y direction. Resolving forces, \[ N = F_g \sin(30) = \frac{1}{2}mg \] and the net force in the y direction is zero. Hence \[ F_f = \mu N = (0.1)\frac{1}{2}mg = 0.05 mg \] and this force is in the negative x direction, decelerating the sled. Now that means the equations of motion in the x direction are - for velocity in the x-direction: \[ v(t) = 15 - 0.05mgt \] - for dispalcement in the x-direction: \[ x(t) = 15t - \frac{0.05}{2}gt^2 \] Now you should be able to solve your problem. The sled reaches as far as it reaches when the velocity v(t) = 0. Use the value of t you find that from that equation to find the corresponding x(t)
JamesJ
  • JamesJ
**correction: the calculation of the normal force above with sin isn't right. It should be \[ N = F_g \cos(30) \] Apologies. You'll need to track that change through the problem.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.