A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 4 years ago

A 32.4 kg wagon is towed up a hill inclined at 18.4◦ with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 112 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. The acceleration of gravity is 9.8 m/s^2. How fast is the wagon going after moving 43.8 m up the hill? Answer in units of m/s

  • This Question is Closed
  1. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1


  2. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1327726001389:dw|By paying attention to the geometry of our situation we see that the force of gravity acting against the tension of the rope is\[|F_g|\sin\theta\]The total force acting on the object along the direction of the incline will be proportional to its acceleration, which we can find by summing the forces along the side of the hill..\[F=T-|F_g|\sin\theta\]\[ma=T-mg\sin\theta\]Assuming 'up the hill' is in the direction along the theta direction we can calculate final velocity from the kinematic equation\[v_f^2=v_o^2+2ad\]In our case vo=0 because the wagon starts from rest.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.