Answer:
a) 10.0 m/s
b) -4.68 m/s
Explanation:
Given:
y₀ = 0 m
y = 4.00 m
t = 1.50 s
a = -9.8 m/s²
Find: v₀, v
y = y₀ + v₀ t + ½ at²
4.00 = 0 + v₀ (1.50) + ½ (-9.8) (1.50)²
v₀ = 10.0 m/s
v = at + v₀
v = (-9.8) (1.50) + 10.0
v = -4.68 m/s
Answer:
14.14 m/s
Explanation:
As total mechanical energy is conserved, if the potential energy is the same at Chad's position no matter if the ball is travelling up or down, then its kinetic energy, and speed, is also the same too.
Therefore, the ball would have a speed of 10m/s when it's passing Chad the 1st time. Since we know that Chad is at 5m high from the release point, we can use the following equation of motion:
where v = 10 m/s is the velocity of the ball when it passes Chad, v_0 is the initial velocity of the ball when it releases, g = 10 m/s2 is the deceleration of the can, and is the distance traveled between Nicole and Chad. We can solve for v0:
The magnitude is the force pushed to make an acceleration
Answer:
An 8 kg sled mass moves in a straight line on a surface without horizontal friction at a certain point its speed is 4 m / s 2.50 m later its speed is 6 m / s using the work and energy theorem, determine the force acting on the sled assuming it is constant and acts in the direction of the sled movement
The force acting on the 8 kg sled has a value of 32 N.
Explanation:
We have initially look for the work that is applied to the sled, and this will be nothing more than the difference in kinetic energy from the end point to the initial, then:
ΔEc = Ec₂ - Ec₁
ΔEc = 0.5 * m * [V₂² - V₁²]
ΔEc = 0.5* 8 kg * [(6m / s) ² - (4m / s) ²]
ΔEc = 80 J
So, the work is 80 J, now we know that work is a relation distance and force , such that:
W = F * d
F = W / d
F = (80 J) / (2.50 m)
F = 32 N
Therefore, the force acting on the sled is 32 N.