Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U =
................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 = 
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U = 
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J
Answer:
6.746 ft/s^2
Explanation:
v(t)=50
v(0)=27
t=5/3600 = 1/720 hours
v(t)-v(0)= a(t-0)
50-27= a(1/720)
a= 23*720= 16560 mi/h^2
16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2
Answer:
a) 
b) 
Explanation:
Given:
- initial rotational speed of phonograph,

- final rotational speed of phonograph,

- time taken for the acceleration,

a)
Now angular acceleration:



b)
Using eq. of motion:



F=ma
As velocity is constant, a=0
So, F=0
Hope this helps!