- 42,802
- 10,495
I get slightly less, but that should be close enough. Maybe you used a larger value for g.DeepPatel said:I had added 880J and I think I had to subtract. If I fixed the errors I now got an answer of 20.549 m/s
The discussion focuses on calculating the work done by friction and the final velocity of a man sledding down a 20m high hill at a 30-degree angle, starting with an initial speed of 7m/s and experiencing a friction force of 22N. The work done by friction was calculated using the equation W=Fdcos(θ), resulting in 880J, which was then used to determine the final velocity. The final speed at the bottom of the hill, after accounting for the negative work done by friction, was found to be approximately 20.549 m/s.
PREREQUISITESStudents studying physics, particularly those focusing on mechanics, as well as educators looking for examples of energy conservation and friction in real-world applications.
I get slightly less, but that should be close enough. Maybe you used a larger value for g.DeepPatel said:I had added 880J and I think I had to subtract. If I fixed the errors I now got an answer of 20.549 m/s
Yeah our teacher prefers we use 10 m/s2 instead of 9.81m/s2 to save us time and make it easier.haruspex said:I get slightly less, but that should be close enough. Maybe you used a larger value for g.