Respuesta :
The total distance travel by the object is 2.5(F0/m)(t0)^2
Since constant force is applied on the object so we can apply the equation of linear motion.
According to law of inertia
F=ma
therfore for the time period t=0 to t_o the acceleration of the object is
a=F0/m
And the velocity of the object after t0 sec is calculated by the first equation of motion
v=u+at
since the initial velocity of the object is 0, therefore the velocity of the object after t0 sec is
v=(F0/m)*t0
and the distance cover by the object during t0 sec is calculated by the second equation of motion
s1=ut+0.5at^2
Plugging the values
s1=0.5(F0/m)(t0)^2
Now the acceleration of the object during 2t0 sec is
a2=2F0/m
therefore the distance travel by the object from time t0 to 2t0 is calculated as
s2=(F0/m)*t0*t0+0.5(2F0/m)*(t0)^2
=(2F0/m)(t0)^2
Therefore total distance travel by the object is
S=s1+S2
=2.5(F0/m)(t0)^2
[(5F₀)/(2m)] t₀²
Further explanation
We deal with a case problem of uniformly accelerated motion and Newton's Second Law.
The formulas used are given by
[tex]\boxed{ \ F = ma \ } \ \boxed{ \ v = u + at \ } \ \boxed{ \ s = ut + \frac{1}{2}at^2 \ }[/tex]
- F = force
- m = mass
- a = acceleration
- u = initial velocity
- v = final velocity
- s = distance travelled
- t = time taken
The first condition from t = 0 to t = t₀
A mass m is at rest on a horizontal frictionless surface at t = 0, thus the initial velocity [tex]\boxed{ \ u = 0 \ }.[/tex]
A constant force f₀ acts on it for a time t₀.
We call the acceleration in the first condition as a₁, i.e.,
[tex]\boxed{ \ a_1 = \frac{F_1}{m} \ } \rightarrow \boxed{ \ a_1 = \frac{F_0}{m} \ } \ (Equation-1)[/tex]
The final velocity at t = t₀ is given by
[tex]\boxed{ \ v_1 = 0 + a_1t_1 \ } \rightarrow \boxed{ \ v_1 = \Big( \frac{F_0}{m} \Big) t_0 \ } \ (Equation-2)[/tex]
The distance travelled from t = 0 to t = t₀ is given by
[tex]\boxed{ \ x_1 = (0)t + \frac{1}{2}a_1t_1^2 \ } \rightarrow \boxed{ \ x_1 = \frac{1}{2} \Big(\frac{F_0}{m} \Big)t_0^2 \ } \ (Equation-3)[/tex]
The second condition from t = t₀ to t = 2t₀
A constant force 2f₀ acts on it from t = t₀ to t = 2t₀.
We call the acceleration in the second condition as a₂.
[tex]\boxed{ \ a_2 = \frac{F_2}{m} \ } \rightarrow \boxed{ \ a_2 = \frac{2F_0}{m} \ } \ (Equation-4)[/tex]
The initial velocity of the second condition is precisely the final velocity of the first condition. We call back Equation-2.
[tex]\boxed{ \ u_2 = v_1 = \Big( \frac{F_0}{m} \Big) t_0 \ } \ (Equation-5)[/tex]
Time taken from t = t₀ to t = 2t₀ is [tex]\boxed{ \ t_2 = 2t_0 - t_0 \rightarrow t_2 = t_0 \ } \ Equation-6[/tex]
The distance travelled from t = 0 to t = t₀ is given by
[tex]\boxed{ \ x_2 = u_2t_2 + \frac{1}{2}a_2t_2^2 \ } \ (Equation-7)[/tex]
Substitute Equation-4, Equation-5, and Equation-6 into Equation-7.
[tex]\boxed{ \ x_2 = \Big( \frac{F_0}{m} \Big) t_0 \cdot t_0 + \frac{1}{2} \Big(\frac{2F_0}{m} \Big) t_0^2 \ }[/tex]
[tex]\boxed{ \ x_2 = \Big( \frac{F_0}{m} \Big) t_0^2 + \Big(\frac{F_0}{m} \Big) t_0^2 \ }[/tex]
[tex]\boxed{ \ x_2 = 2 \Big( \frac{F_0}{m} \Big) t_0^2 \ } \ (Equation-8)[/tex]
Determine the total distance
The total distance travelled from t = 0 to t = 2t₀ is obtained from the sum of Equation-3 and Equation-8.
[tex] \boxed{ \ x_{total} = x_1 + x_2 \ } [/tex]
[tex]\boxed{ \ x_{total} = \frac{1}{2} \Big(\frac{F_0}{m} \Big)t_0^2 + 2 \Big( \frac{F_0}{m} \Big) t_0^2 \ }[/tex]
[tex]\boxed{ \ x_{total} = \frac{1}{2} \Big(\frac{F_0}{m} \Big)t_0^2 + \frac{4}{2} \Big( \frac{F_0}{m} \Big) t_0^2 \ }[/tex]
Thus, the total distance travelled is [tex]\boxed{\boxed{ \ x_{total} = \frac{5}{2} \Big(\frac{F_0}{m} \Big) t_0^2 \ }}[/tex]
Learn more
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Keywords: uniformly accelerated motion, Newton's second law, A mass, at rest on a horizontal frictionless, a constant force, for a time. suddenly the force doubles, determine the total distance travelled
