Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
Answer:
positive
Explanation:
The ball is rolling down with a negative velocity, but the velocity is slowing down. therefore the velocity must increase in order for the ball to slow down.
For example let the ball's initial velocity be -15 m/s. and it is slowing down to let's say -13 m/s. Well this means that it's velocity has increase by 2 m/s. So, its acceleration is positive.
Answer:
In an elastic collision:
- There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.
- Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.
- There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.
- Velocities may change after collision. If the masses are equal, the velocities interchange.
When one object is stationary:
Final velocity of object 1:
v₁ = (m₁ - m₂)u₁/(m₁ +m₂)
Final velocity of object 2:
v₂ = (2 m₁ u₁)/(m₁+m₂) =
- Objects do not stick together in elastic collision. They stick together in inelastic collision.
- One object may be stationary before the elastic collision.
Thus, conditions for an elastic collision:
- Energy is conserved.
- Velocities may change.
- Momentum is conserved.
- Kinetic energy is conserved.
- One object may be stationary before the elastic collision.
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction =
=
= 0.12983
The average coefficient of static friction is 0.130