m = mass of the ball being raised
h = height to which the ball is raised = 0.5 m
v₀ = initial speed given to the ball = 3
v = final speed of the ball at its lowest position = ?
using conservation of energy
final kinetic energy at the lowest point = initial kinetic energy + initial potential energy
(0.5) m v² = (0.5) m v₀² + mgh
dividing each term by "m"
(0.5) v² = (0.5) v₀² + gh
inserting the values
(0.5) v² = (0.5) (3)² + (9.8) (0.5)
v = 4.34 m/s
Answer:
No car will slide
Explanation:
Neither car will slide because proper bank does not required frictional forces. To prevent the car from traction through curve, a component of normal forces provides necessary centripetal forces. which prevent them to slide.
No car will slide since both car are going with the proper speed for given banking angle
Answer:
B. Kinetic energy.............
Answer:
Ki = 0.28665 J
h = 0.133 m
Explanation:
Given:
- The mass of rod M = 0.22 kg
- The length of rod L = 1.2 m
- The angular speed at the lowest point w = 2.33 rad /s
Find:
(a) the rod's kinetic energy at its lowest position
(b) how far above that position the center of mass rises.
Solution:
- The moment of inertia of a rod pivoted at one of its ends is given by I:
I = ML^2 / 3
- The Kinetic energy at the lowest point is given by the rotational energy as follows:
Ki = 0.5*I*w^2
Ki = 0.5*ML^2 / 3*w^2
Ki = ML^2*w^2 / 6
Ki = (.22*1.2^2*2.33^2) / 6
Ki = 0.28665 J
- Since no external force was acting on the rod we can apply the conservation of energy of system consisting of the rod where the change in kinetic energy leads to a change in gravitational potential Energy:
Kf - Ki = Pf - Pi
0 - 0.28665 J = mg( 0 - h )
mg*h = 0.28665
h = 0.28665 / ( 0.22*9.81 )
h = 0.133 m