An object of mass m = 200 g slides down a frictionless ramp from a height H = 60 cm. Near the bottom of the ramp, the path changes into an arc of a circle with radius r = 20 cm. The arc has coefficient of kinetic friction $\mu$_k = 0.4. The angle $\theta$ = 120 $\degree$ and points P₁ and P₂ which are the ends of the arc, are at the same height. What is the speed of the object at point P₁? A) 1.1 m/s B) 2.5 m/s C) 3.1 m/s D) 3.4 m/s E) 4.1 m/s

Question

Quizplus · Accepted Answer

First, we need to find the velocity of the object at the bottom of the ramp before it enters the circular path. We can use the conservation of energy:

mgh = 1/2mv^2
v = sqrt(2gh)

where m = 200 g = 0.2 kg, g = 9.8 m/s^2, and h = 60 cm = 0.6 m

v = sqrt(2 * 9.8 m/s^2 * 0.6 m) ≈ 3.14 m/s

Next, we can use the fact that the object is moving in a circular path with a radius of 20 cm = 0.2 m and a coefficient of kinetic friction of 0.4 to find the maximum speed it can have at the bottom of the arc without slipping:

μk = v^2 / rg
vmax = sqrt(μk * rg)

where μk = 0.4, r = 0.2 m

vmax = sqrt(0.4 * 9.8 m/s^2 * 0.2 m) ≈ 1.1 m/s

Finally, we can use the conservation of energy again to find the speed of the object at the end of the arc (point P1):

1/2mv^2 = 1/2mvmax^2 + (μk * mg * d)
where d = r(1 - cosθ) = 0.2 m(1 - cos 120°) ≈ 0.4 m

v = sqrt(vmax^2 + 2(μk*g*d))

v ≈ sqrt((1.1 m/s)^2 + 2(0.4)(9.8 m/s^2)(0.4 m)) ≈ 3.1 m/s

Therefore, the best choice is (C).

An Object of Mass M = 200 G Slides Down μ\muμ

An Object of Mass M = 200 G Slides Down $\mu$