Respuesta :
Answer:
F = 51 10² N and The force applied is 11.6 times greater than the weight
Explanation:
As we are asked the force we can use Newton's second law and we calculate the acceleration by kinematics
y = v₀ t + ½ a t²
On the ground its initial velocity is zero (v₀ = 0)
y = ½ a t²
a = 2y / t²
a = 2 2.5 / 0.21²
a = 113.38 m / s²
We calculate the force
F = ma
F = 45 113.38
F = 5102 N
F = 51 10² N
For the relationship of this force with the weight, divide the two
F / W = 51 10²/45 9.8
F / W = 11.6
The force applied is 11.6 times greater than the weight
Answer:
[tex]F = 5100 N[/tex]
[tex]R = 11.56[/tex]
Explanation:
As we know that impala is pushed upwards straight up with uniform acceleration
So here we can use kinematics equation
So we will have
[tex]y = vt + \frac{1}{2}at^2[/tex]
[tex]2.5 = \frac{1}{2}a(0.21)^2[/tex]
[tex]a = 113.4 m/s^2[/tex]
now we have
[tex]F = ma[/tex]
[tex]F = 45(113.4)[/tex]
[tex]F = 5100 N[/tex]
Now ratio of this force with the weight is given as
[tex]R = \frac{F}{F_g}[/tex]
[tex]R = \frac{5100}{45 \times 9.81}[/tex]
[tex]R = 11.56[/tex]
