A33.1 g copper object is launched from a 1.5 m 30° steel incline positioned on the floor by being pulled up a string attached to a 50.0 g mass suspended vertically over a pulley. the object is projected towards a glass table where it lands when it is at the point along its trajectory with the lowest speed. it comes to a halt when it clears the opposite edge of the table. it then falls and lands on 9 physics texts each 5 cm thick that are stacked on the floor on the opposite side of the table. assume that the table does not have a ledge and the rectangular object experiences 0.05 n of air resistance as it falls towards the books. how far and how long did the object travel and how fast does it hit the books. assume that at the moment the copper object leaves the incline, the massless string and ideal pulley break off

Explanation:

To answer how far and how long the object travels, we need to know either the length of the table or the kinetic coefficient of friction between copper and glass.  We also need to know the kinetic coefficient of friction between the copper object and the steel incline.  This information wasn't provided, so for sake of illustration, I'll assume both coefficients are the same, and that μk = 0.35.

Let's divide the path into four sections to stay organized:

a) object moves up incline

b) object is in free fall

c) object slides across table

d) object falls onto books

a)

First, the object slides up the incline.  There are four forces acting on the object.  Weight pulling down, normal force perpendicular to the incline, tension up the incline, and friction down the incline.

Sum of the forces normal to the incline:

∑F = ma

N - W cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

T - F = ma

T - Nμ = ma

T - mgμ cos θ = ma

Now sum of the forces in the y direction on the hanging mass:

∑F = ma

T - W = M(-a)

T = Mg - Ma

Substituting:

Mg - Ma - mgμ cos θ = ma

Mg - mgμ cos θ = (m + M) a

a = g (M - mμ cos θ) / (m + M)

Given m = 33.1 g, M = 50.0 g, θ = 30°, μ = 0.35, and g = 9.8 m/s²:

a = 4.71 m/s²

So the velocity it reaches at the top of the incline is:

v² = v₀² + 2a(x - x₀)

v² = 0² + 2(4.71)(1.5)

v = 3.76 m/s

The time to reach the top of the incline is:

x = x₀ + v₀ t + ½ at²

1.5 = 0 + (0) t + ½ (4.71) t²

t = 0.798 s

The horizontal distance traveled is:

x = 1.5 cos 30°

x = 1.299 m

And the vertical distance traveled is:

y = 1.5 sin 30°

y = 0.750 m

b)

In the second stage, the object is in free fall.  v₀ = 3.76 m/s and θ = 30°.  The object lands on the table at the point where its speed is a minimum.  This is at the highest point of the trajectory, when the vertical velocity is 0.

v = at + v₀

0 = (-9.8) t + (3.76 sin 30°)

t = 0.192 s

The horizontal distance traveled is:

x = x₀ + v₀ t + ½ at²

x = 0 + (3.76 cos 30°) (0.192) + ½ (0) (0.192)²

x = 0.625 m

The height it reaches is:

v² = v₀² + 2a(y - y₀)

(0)² = (3.76 sin 30°)² + 2(-9.8)(y - 0)

y = 0.180 m

c)

The object is now on the glass table.  As it slides, there are three forces acting on it.  Normal force up, gravity down, and friction to the left.

In the y direction:

∑F = ma

N - W = 0

N = mg

In the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

The time it takes to reach the end of the table is:

v = at + v₀

0 = (-9.8×0.35) t + (3.76 cos 30°)

t = 0.949 s

And the distance it travels is:

v² = v₀² + 2a(x - x₀)

(0)² = (3.76 cos 30°)² + 2(-9.8×0.35) (x - 0)

x = 1.546 m

d)

Finally, the object falls onto the books.  There are 9 books, each 5 cm thick, so the height is 45 cm.  As the book falls, there are two forces acting on it: drag up and gravity down.

∑F = ma

D - W = ma

0.05 - (0.0331)(9.8) = 0.0331 a

a = -8.29 m/s²

Adding the vertical distances from parts a and b, we know the height of the table is 0.930 m.  So the time it takes the object to land is:

y = y₀ + v₀ t + ½ at²

0.45 = 0.93 + (0) t + ½ (-9.8) t²

t = 0.313 s

Adding up the times from all four parts, the total time is:

t = 0.798 + 0.192 + 0.949 + 0.313

t = 2.25 s

Adding up the horizontal distances, the total distance traveled is:

x = 1.299 + 0.625 + 1.546

x = 3.47 m

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2