Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)
• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)
•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
Answer:d
Explanation:
Spring is compressed to a distance of x from its equilibrium position
Work done by block on the spring is equal to change in elastic potential energy
i.e. Work done by block 
therefore spring will also done an equal opposite amount of work on the block in the absence of external force
Thus work done by spring on the block 
Thus option d is correct
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Explanation:
Part a)
Angular speed is given as



Part b)
Since turn table is accelerating uniformly
so we will have




Part c)
angular acceleration is given as



Part d)
When its angular speed changes to 120 rpm
then we will have


number of turns revolved is 15 times
so we have



Part e)
now for uniform acceleration we have



r = radius of the circle of the ride = 3.00 meters
v = linear speed of the person during the ride = 17.0 m/s
m = mass of the person in angular motion in the ride
L = angular momentum of the person in the ride = 3570 kg m²/s
Angular momentum is given as
L = m v r
inserting the values
3570 kg m²/s = m (17 m/s) (3.00 m)
m = 3570 kg m²/s/(51 m²/s)
m = 7 kg
hence the mass comes out to be 7 kg