A moving car initially has kinetic energy K. The car then moves in the opposite direction with four times its initial speed. Wha
1 answer:
You might be interested in
(a) 34.6 N
To solve the problem, we have to analyze the forces acting along the horizontal direction.
We have:
- Forward: the component of the pull parallel to the ground, which is
where
F is the magnitude of the pull
is the angle
- Backward: the force of friction, which is
So, the equation of motion is
where
m = 20 kg is the mass of the wagon
a is the acceleration
In this part, the wagon is moving at constant speed, so a =0 and the equation becomes
Therefore, we can find the pulling force:
(b) 43.9 N
In this case, the acceleration is
So, the equation of motion in this case is
So this time we have to take into account the term (ma).
Using the same data as before:
m = 20 kg
We find the new magnitude of F:
Based on the forces acting on the axes, the resultant moments will be (345, 400, 600 N·m)
<h3>What would be resultant moment about x-axis?</h3>= F₃ x 3
= -115 x 3
= -345 N·m
<h3>What would be resultant moment about y-axis?</h3>= F₁ x 2
= -200 x 2
= -400 N·m
<h3>What would be the resultant moment about z-axis?</h3>= F₄ x 2
= -300 x 2
= - 600 N·m
In conclusion, the resultant moment about x, y, and z axes is (345, 400, 600 N·m)
Find out more on resultant moments at brainly.com/question/6278006.
