In particle kinematics, we measure the acceleration of the particle. For this, an observer is taken as a reference point. Generally we consider the observer as in origin of a co-ordinate system. The place where observer is placed is known as frame of reference.
Newton's laws only work in the inertial frame of reference. Inertial frame is a frame of reference whose acceleration is zero. But, if you are observing the motion of an object from non-inertial frame, you need to add and extra force on the system under observation which is known as pseudo force.
Answer:
Your little sister (mass 25.0 ) is sitting in her little red wagon (mass 8.50 at rest. You begin pulling her forward and continue accelerating her with a constant force for 2.35 at the end of which time she's moving at a speed of 1.80 .
(a) Calculate the impulse you imparted to the wagon and its passenger. (b) With what force did you pull on the wagon?
1) Call F1 the larger force and F1x and F1y its its x-and-y- components.respectively.
I will use the complementary angle: 90 - 25 = 65 to work with the normal convention.
=> cos(65) = F1x / F1 => F1x = - F1*cos(65) (I choose negative as the west direction)
=> sin(65) = F1y / F1 => F1y = F1*sin(65) (I choose positive the north direction)
2) Call F2 the shorter force and F2x and F2y its components
=> cos(x) = F2x / F2 => F2x = F2*cos(x)
=> sin(x) = F2y / F2=> F2y = F2*sin(x)
3) You know that:
- F1 = 2F2
- The net force in the y direction is 430 N
- The net force in the x direction is 0
a) F1x + F2x = 0
=> -F1*cos(65) + F2*sin(x) = 0
=> F1*cos(65) = F2 sin(x) => sin(x) = [F1/F2] cos(65)
Remember F1 = 2F2 => F1/F2 = 2 => sin(x) = 2 cos(65) = 0.84524
=> x = arcsin(0.84524) = 57.7
b) F1y + F2y = 430 =>
F1 sin(65) + F2*sin(57.7) = 430 =>
0.9060F1 + 0.84524F2 430
F1 = 2F2 => 0.9060*2F2 + 0.84524F2 = 430 => 1.7512F2 = 430
=> F2 = 430 / 1.7512 = 245.54 N
=> F1 = 2*245.54 =491.1N
There you have the two forces.
The angle of the shorter force is 57.7 measured from the east to the north (this is north of east), which would be 90 - 57.7 = 32.3 degrees east of north..
Then the shorter force is 245.5 N at 32.3 degrees east of north
And the larger force is 491.1 N at 25.0 degrees west of north.
Answer:
k = 22.05 N/m
Explanation:
The potential energy of the mass is converted into potential energy of the spring.
Given:
mass m = 0.27 kg
gravitational constant g = 9.8 m/s²
distance falling/ stretching of spring h = 0.24 m
Solving for k:
