A 2kg body is pushed by a force of 35N a distance of 3m up a ramp that is inclin

Question

A 2kg body is pushed by a force of 35N a distance of 3m up a ramp that is inclined 40 degrees. Force P is parallel to ramp. I found that there is 80.4J of work done by force P, and Gravitational Potential energy of the body was 37.8J. Now the body gets stopped at the top of ramp and then released. It slides down the ramp abainst a friction force of 10N. Find the body's change in kinetic energy in Joules after traveling the 3m down the ramp. I thought KE(initial) + PE(initial)= KE(Final) + PE(final). But I cannot seem to get it to work out for me. Can anyone lead me in the right direction? Oh I did find that y at the top of ramp is 1.93m.

Explanation / Answer

Since the potential energy at the top of the ramp is 37.8 J,that would be the kinetic energy at the bottom if there were nofriction. However, there is, so we need to subtract off thework done by friction. Since you have so graciouslycalculated that, I don't need to. : ) You said the work done by friction was 30 J. So, 37.8 -30 = 7.8 J of kinetic energy at the bottom of the ramp. Edit: Oh and there's 0 kinetic energy at the top of theramp, so the change would be + 7.8 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.