Question

Определите работу силы трения, если масса конькобежца 50 кг и продвижении по инерции до остановки он за 10 секунд проезжает 10 м , двигаясь равнозамедленно.

Answer 1

Дано    m=50 кг   t=10c    V=0    S=10м     А- ?

S=(Vo+V)*t/2
Vo=2*S/t - начальная скорость
Vo=2*10/10=2 м/с
m*Vo^2/2 + Атр=0
Атр=-m*Vo^2/2=-50*4/2=-100 Дж

Answer 2

Understanding the Problem

The problem states that a skater with a mass of 50 kg comes to a stop after traveling 10 meters in 10 seconds. The skater is moving with uniform deceleration. We need to determine the force of friction acting on the skater.

Solution

To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the force of friction.

We can start by calculating the acceleration of the skater using the formula:

acceleration = (final velocity - initial velocity) / time

Since the skater comes to a stop, the final velocity is 0 m/s. The initial velocity can be calculated using the formula:

initial velocity = distance / time

Substituting the given values, we have:

initial velocity = 10 m / 10 s = 1 m/s

Now we can calculate the acceleration:

acceleration = (0 m/s - 1 m/s) / 10 s = -0.1 m/s^2

The negative sign indicates that the skater is decelerating.

Next, we can calculate the force of friction using the formula:

force of friction = mass * acceleration

Substituting the given values, we have:

force of friction = 50 kg * (-0.1 m/s^2) = -5 N

The negative sign indicates that the force of friction is acting in the opposite direction of the skater's motion.

Therefore, the force of friction acting on the skater is 5 N in the opposite direction of the skater's motion.