Answer:
K' = 1777.777 J
Explanation:
Given that
m = 40 kg
v= 15 m/s
K=1000
Given that kinetic energy(K) varies with mass(m) and velocity(v)
K= C(mv²)
Where
C= Constant
m=mass
v=velocity
When
m = 40 kg ,v= 15 m/s ,K=1000
K= C(mv²)
1000 = C( 40 x 15²)
C=0.111111
When m = 40 kg and v= 20 m/s
K' = C(mv²)
K= 0.1111 x (40 x 20²)
K' = 1777.777 J
complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
Explanation:
Take north to be positive and south to be negative.
a = (v − v₀) / t
a = (-4.5 m/s − 4.5 m/s) / 8 s
a = -1.125 m/s²
The acceleration is 1.125 m/s² south.
