Answer:
a = 0.009 J
b = 0.19 m/s
c = 0.005 J and 0.004 J
Explanation:
Given that
Mass of the object, m = 0.5 kg
Spring constant of the spring, k = 20 N/m
Amplitude of the motion, A = 3 cm = 0.03 m
Displacement of the system, x = 2 cm = 0.02 m
a
Total energy of the system, E =
E = 1/2 * k * A²
E = 1/2 * 20 * 0.03²
E = 10 * 0.0009
E = 0.009 J
b
E = 1/2 * k * A² = 1/2 * m * v(max)²
1/2 * m * v(max)² = 0.009
1/2 * 0.5 * v(max)² = 0.009
v(max)² = 0.009 * 2/0.5
v(max)² = 0.018 / 0.5
v(max)² = 0.036
v(max) = √0.036
v(max) = 0.19 m/s
c
V = ±√[(k/m) * (A² - x²)]
V = ±√[(20/0.5) * (0.03² - 0.02²)]
V = ±√(40 * 0.0005)
V = ±√0.02
V = ±0.141 m/s
Kinetic Energy, K = 1/2 * m * v²
K = 1/2 * 0.5 * 0.141²
K = 1/4 * 0.02
K = 0.005 J
Potential Energy, P = 1/2 * k * x²
P = 1/2 * 20 * 0.02²
P = 10 * 0.0004
P = 0.004 J
Answer:
20 J
Explanation:
Given:
Weight of the book is, 
Height or displacement of the book is, 
The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.
So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.
Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.
Therefore, work done is given as:

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:

where
m=225 kg is the mass of the boat

is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:

and the negative sign means the direction of the impulse is against the direction of motion of the boat.