The ball X has a known position at 3 different times, each 1 second apart. The positions of the ball are (1.8,2.2)m, (4.4,4.8)m,
1 answer:
Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x =
Similarly velocity along y direction V_y(1) =
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²
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To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:
Where,
F = Force
r = Radius
Replacing we have that,
The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore
Finally, angular acceleration is a result of the expression of torque by inertia, therefore
PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians , therefore
Answer:
θ = 13.7º
Explanation:
- According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
- The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
- As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:
- ΔK, is the change in kinetic energy, as follows:
- ΔU, is the change in the gravitational potential energy.
- If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:
- Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:
- Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:
- Replacing by the givens and the knowns, we can solve for sin θ, as follows:
⇒ θ = sin⁻¹ (0.236) = 13.7º