A 200 gram mass moves on a horizontal friction surface and itsposition vector is

Question

A 200 gram mass moves on a horizontal friction surface and itsposition vector is given in SI units by: vector r = 2.50t^2 x-hat -3.75t^3 y-hat. At t = 0.500s, what is the magnitude of the net, externalforce action on the mass? heres what i got so far, im not sure if its right. a(t) = 5.00x-hat - 22.5t y-hat. a(t) = -6.25. So force net is .2(-6.25) =1.25 Is that right? And it also says what is the direction relativeto the positive x axis of the net external force? Im not sure howto do that. Thanks in advance A 200 gram mass moves on a horizontal friction surface and itsposition vector is given in SI units by: vector r = 2.50t^2 x-hat -3.75t^3 y-hat. At t = 0.500s, what is the magnitude of the net, externalforce action on the mass? heres what i got so far, im not sure if its right. a(t) = 5.00x-hat - 22.5t y-hat. a(t) = -6.25. So force net is .2(-6.25) =1.25 Is that right? And it also says what is the direction relativeto the positive x axis of the net external force? Im not sure howto do that. Thanks in advance

Explanation / Answer

r = 2.50t2 i - 3.75t3 j ==> v = 5t i - 11.25t2 j ==> a = 5i - 22.5t j Therefore F = ma = 0.2 * [5i - 22.5t j] = i - 4.5j magnitude = 4.61 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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