Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;

(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for
, we find:

So, the momentum of block B after the collision is -50 kg m/s.
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Answer:
Coefficient of friction will be 0.296
Explanation:
We have given initial speed of the stone u = 8 m /sec
It comes to rest so final speed v = 0 m /sec
Distance traveled before coming to rest s = 11 m
According to third equation of motion

So 

Acceleration due to gravity 
We know that acceleration is given by

So 

So coefficient of friction will be 0.296