31.3m/s
Explanation:
Given parameters:
Mass of rock = 40kg
Height of cliff = 50m
Unknown:
Speed of rock when it hits ground = ?
Solution:
We are going to use the appropriate motion equation to solve this problem
The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.
Using;
V² = U² + 2gH
V = unknown velocity
U = initial velocity = O
g = acceleration due to gravity = 9.8m/s²
H = height of fall
since the initial velocity of the bodyg is 0
V² = 2gH
V= √2gH = √2 x 9.8 x 50 = 31.3m/s
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Explanation:
after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s
The displacement after 5s is
x=vo + (1/2)at²
x = 0 + (1/2)(3m/s²)(5s)(5s)
x= 37.5 m
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
Answer:
d = 0.05 [m] = 50 [mm]
Explanation:
We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.