Let us consider a material point with mass m = 4 kg moving along a horizontal straight line. We need to determine after what time the speed of the point will decrease by 10 times, provided that the force of resistance to movement is equal to R = 0.8v.
We use Newton's second law:
$$ F = in $$
where F is the force acting on a material point, m is its mass, and is the acceleration of the point.
The force of resistance to movement R can be expressed in terms of speed v:
$$ R = 0,8v $$
Then the equation of motion of the material point will take the form:
$$ m\frac{dv}{dt} = -R $$
where t is the time elapsed since the beginning of the movement.
Substituting the expression for R, we get:
$$ m\frac{dv}{dt} = -0,8v $$
Dividing both sides of the equation by m and moving the variables, we get:
$$ \frac{dv}{v} = -\frac{0,8}{m}dt $$
Let's integrate this equation from the initial speed v0 to the speed v through time t:
$$ \int_{v_0}^v \frac{dv}{v} = -\frac{0,8}{m} \int_0^t dt $$
After integration we get:
$$ \ln\frac{v}{v_0} = -\frac{0,8}{m}t $$
Let us express v in terms of v0:
$$ v = v_0e^{ -\frac{0,8}{m}t} $$
Now you can find the time after which the speed of the point will decrease by 10 times. To do this, substitute the value v0/10 instead of v into the equation v = v0e^(-0.8t/m):
$$ \frac{v_0}{10} = v_0e^{ -\frac{0,8}{m}t} $$
Dividing both sides by v0 and taking the natural logarithm, we get:
$$ \ln\frac{1}{10} = -\frac{0,8}{m}t $$
From here:
$$ t = \frac{m}{0,8} \ln 10 \approx 11,5 \text{ сек} $$
Thus, after 11.5 seconds the speed of the material point will decrease by 10 times with a force of resistance to motion equal to 0.8v.
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This product contains a detailed solution to the problem associated with the movement of a material point along a horizontal straight line. The problem requires determining the time after which the speed of a point will decrease by 10 times at a given force of resistance to movement.
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Solution to problem 13.2.25 from the collection of Kepe O.?. consists in determining the time after which the speed of a material point weighing 4 kg, moving along a horizontal straight line, will decrease by 10 times at a given motion resistance force R = 0.8v.
In this problem, you can use Newton's second law F = ma, where F is the force acting on a material point, m is its mass, a is acceleration. You can also use the law of change in kinetic energy ΔK = K2 - K1 = W, where K1 and K2 are the initial and final kinetic energies of a material point, respectively, W is the work done by the force of resistance to movement.
First you need to determine the acceleration of a material point. From Newton's second law F = ma we obtain that a = F/m. According to the conditions of the problem, the force of resistance to movement is equal to R = 0.8v, where v is the speed of the material point. Thus, a = 0.8v/m.
Next, you need to determine the time after which the speed of the material point will decrease by 10 times. Let us denote the initial velocity of the material point as v0, and the final velocity as v. Then from the law of change in kinetic energy ΔK = K2 - K1 = W we obtain:
m(v^2 - v0^2)/2 = -RWt,
where t is time, W = -RWt is the work done by the force of resistance to movement.
In order for the speed to decrease by 10 times, it is necessary that v = v0/10. Substituting this value into the equation above and solving it for t, we get:
t = (m/8R) * ln(10)
Substituting the values m = 4 kg and R = 0.8v/m into the formula, we get:
t ≈ 11.5 sec
Thus, the answer to the problem: after 11.5 seconds, the speed of a material point with a mass of 4 kg moving along a horizontal straight line will decrease by 10 times at a given force of resistance to movement R = 0.8v.
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