12 kg wagon is being pulled at an angle of 38o above horizontal. What force is applied to the wagon if it accelerates from rest to a speed of 2.2 m/s over a distance of 3.4 m? Is this question a typo? Am I supposed to solve for force or work?
I need help
F cos 38 = m a
easy way is for constant acceleration
v average = (Vi+Vf)/2
so
v average = (0 + 2.2)/2 = 1.1 m/s
so time = t = 3.4/1.1 = 3.09 s
a = change in v / time
= 2.2 / 3.09
= .712 m/s^2
so
F cos 38 = 12 * .712
F = 10.8 Newtons
This is from one of your previous answers. You can do it with work if you wish or you can do it with F = m a
with work
work done = Force in direction of motion * distance
Force in direction of motion = F cos 38
= .788 F
work done = .788 F * 3.4 meters
= 2.68 F Joules
that is the increase in kinetic energy
(1/2) m v^2 = .5*12 * 2.2^2
= 29 Joules
so
2.68 F = 29
F = 10.8 Newtons
Why did you use kinetic energy is equal to work?
This question seems to be asking for the force applied to the wagon. It is not a typo, as force and work are related concepts in this context.
To calculate the force applied to the wagon, we can use Newton's second law of motion:
F = m * a
Where:
F is the force applied to the wagon,
m is the mass of the wagon, and
a is the acceleration of the wagon.
First, we need to calculate the acceleration of the wagon. We can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v is the final velocity of the wagon,
u is the initial velocity of the wagon (which is zero since it's starting from rest),
a is the acceleration of the wagon, and
s is the distance traveled by the wagon.
Plugging in the given values:
v = 2.2 m/s
u = 0 m/s
s = 3.4 m
We can rewrite the equation as:
a = (v^2 - u^2) / (2s)
Substituting in the values, we have:
a = (2.2^2 - 0^2) / (2 * 3.4)
Simplifying the equation:
a = 4.84 / 6.8
a = 0.712 m/s^2
Now, we have the acceleration of the wagon. To calculate the force, we can plug the values into Newton's second law:
F = m * a
Substituting the given value:
m = 12 kg
a = 0.712 m/s^2
We get:
F = 12 kg * 0.712 m/s^2
F = 8.544 N
Therefore, the force applied to the wagon is approximately 8.544 Newtons.