Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
Explanation:
The given data is as follows.
k = 130 N/m,
= 17 cm = 0.17 m (as 1 m = 100 cm)
mass (m) = 2.8 kg
When the spring is compressed then energy stored in it is as follows.
Energy = 
Now, spring energy gets converted into kinetic energy when the box is launched.
So,
= 
= 

= 1.34
v = 1.15 m/sec
Now,
Frictional force = 
= 
= 4.116 N
Also, Kinetic energy = work done by friction
1.8515 =
d = 0.449 m
Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.
Answer:
<em>a. The rock takes 2.02 seconds to hit the ground</em>
<em>b. The rock lands at 20,2 m from the base of the cliff</em>
Explanation:
Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.
The time taken by the object to hit the ground is calculated by:

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.
a,
The time taken by the rock to reach the ground is:


t = 2.02 s
The rock takes 2.02 seconds to hit the ground
b.
The range is calculated now:

d = 20.2 m
The rock lands at 20,2 m from the base of the cliff