Explanation:
A) Use Hooke's law to find the spring constant.
F = kx
40 N = k (0.4 m)
k = 100 N/m
B) Period of a spring-mass system is:
T = 2π √(m / k)
T = 2π √(2.6 kg / 100 N/m)
T = 1 s
Frequency is the inverse of period.
f = 1 / T
f = 1 Hz
The formula we can use in this case is:
d = v0t + 0.5 at^2
v = at + v0
where,
d = distance travelled
v0 = initial velocity = 0 since at rest
t = time travelled
a = acceleration
v = final velocity when it took off
a. d = 0 + 0.5 * 3 * 30^2
d = 1350 m
b. v = 3 * 30 + 0
<span>v = 90 m/s</span>
Answer:
θ = 13.16 °
Explanation:
Lets take mass of child = m
Initial velocity ,u= 1.1 m/s
Final velocity ,v=3.7 m/s
d= 22.5 m
The force due to gravity along the incline plane = m g sinθ
The friction force = (m g)/5
Now from work power energy
We know that
work done by all forces = change in kinetic energy
( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²
(2 g sinθ - ( 2 g)/5 ) d = v² - u²
take g = 10 m/s²
(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²
20 sinθ - 4 =12.48/22.5
θ = 13.16 °
Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
W = m x a
where W is weight, m is mass, and a is acceleration. Substituting the known values,
40 kg m/s² = (10 kg) x a
Calculating for the value of a from the equation will give us an answer equal to 4.
ANSWER: 4 m/s².
I hope I'm not too late.
GPE = mass * gravity * height
GPE = 2 kg * 9.8 m/s * 10
GPE = 196 Joules