Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
Answer:
Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .
Explanation:
The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.
m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N
For any object thrown upwards where only the force of gravity is acting upon it, uses the following formula for the maximum height attained.
H= v²/2g, where g = 9.81 m/s²
There are two information of velocities are given. However, we use the 20 m/s information because this is the launch velocity. Hence, the solution is as follows:
H = (20 m/s)²/2(9.81 m/s²)
<em>H = 20.4 m</em>
It IS <span>PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J </span>
