A small block is released from rest at the top of a frictionless incline. The di

Question

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.10 m. The vertical distance from the top of the incline to the bottom is 1.93 m. If g = 9.80 m/s2, what is the acceleration of the block as it slides down the incline?
A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.10 m. The vertical distance from the top of the incline to the bottom is 1.93 m. If g = 9.80 m/s2, what is the acceleration of the block as it slides down the incline?
g = 9.80 m/s2,

Explanation / Answer

y = 1.93

l = 3.1

sintheta = y/l

gavitational force along the plane = Fg = m*g*sintheta

from newtons law F = m*a

m*a = m*g*sintheta

a = g*sintheta

a = 9.8*1.93/3.1

a = 6.1 m/s^2 <<<_--answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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