Question

Брусок скользит по наклонной поверхности равноускоренно из состояния покоя За вторую секунду движения он прошел путь 60 см Какой путь был пройден бруском за первую секунду движения

Answer 1

Пусть а - ускорение груза. тогда скорсть груза через первую секунду движения V1 = at1. Перемещение груза: s1 = аt1квад/2. Перемещение груза за вторую секунду: s2 = V1t2 + at2квад/2 = at1t2 + at2квад/2. Но t1 = t2 = 1 сек. Получим алгебраическое уравнение для а: а + а/2 = 0,6.  Отсюда а = 0,4 м/с квад.  Теперь находим s1 = at1квад/2 = 0,4/2 = 0,2 м = 20 см.

Answer 2

Calculation of the Distance Traveled by the Block in the First Second

To calculate the distance traveled by the block in the first second of its motion, we need to use the given information that the block slides down a inclined surface with constant acceleration and covers a distance of 60 cm in the second second.

Let's assume that the distance traveled by the block in the first second is represented by S1.

We can use the equation of motion for uniformly accelerated linear motion to find the value of S1. The equation is:

S = ut + (1/2)at^2

Where: - S is the distance traveled - u is the initial velocity (which is zero in this case as the block starts from rest) - t is the time - a is the acceleration

Since the block starts from rest, the initial velocity u is zero. We are given that the motion is uniformly accelerated, so the acceleration a is constant.

Let's assume the acceleration of the block is represented by a.

For the second second of motion, we know that the distance traveled is 60 cm. Using the equation of motion, we can write:

S = ut + (1/2)at^2

Substituting the known values: - S = 60 cm - u = 0 - t = 2 s

We can solve for a:

60 = 0 + (1/2)a(2^2)

Simplifying the equation:

60 = 2a

a = 30 cm/s^2

Now that we have the value of a, we can calculate the distance traveled by the block in the first second using the same equation of motion:

S1 = ut + (1/2)at^2

Substituting the known values: - u = 0 - t = 1 s - a = 30 cm/s^2

We can solve for S1:

S1 = 0 + (1/2)(30)(1^2)

Simplifying the equation:

S1 = 15 cm

Therefore, the block traveled a distance of 15 cm in the first second of its motion.

Answer:

The block traveled a distance of 15 cm in the first second of its motion.