Speed=30 m/s - 1.5 m/s = 28.5 m/s forward
Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm =
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
I would say your answer is B, since Newton's 3rd law is, "For every action, there is an equal and opposite reaction."
It's talking about pairs of actions. Sorry if I'm wrong.