Answer:
117.72 N
Explanation:
The given parameters are;
The mass m₁ = 2.0 × 10³ kg
The mass m₂ = 4.4 × 10² kg
The mass of the man, m₃ = 6.0 × 10 kg
The condition of the interaction of the surfaces = Frictionless surfaces
The
The tension in the string = The downward force = The weight of (m₂ + m₃) = (m₂ + m₃) × g
Let <em>a</em> represent the acceleration of the connected masses due to the weight of m₂, and m₃, we have;
(m₁ + m₂ + m₃) × a = (m₂ + m₃) × g
∴ a = (m₂ + m₃) × g/(m₁ + m₂ + m₃)
Which gives;
a = (4.4 × 10²+ 6.0 × 10) × 9.81/(2.0 × 10³+ 4.4 × 10²+ 6.0 × 10) = 1.962
The downward acceleration, a = 1.962 m/s²
The apparent weight of the man = The mass of the man, m₃ × The acceleration, <em>a</em>
∴ The apparent weight of the man = 6.0×10 kg ×1.962 m/s² = 117.72 N
Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have
Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have
The block's speed is 31.422 cm/s
Answer:
Explanation:
From work energy theorem
Work done by all forces = Change in kinetic energy
Lets take
m= mass of object
h=height from the ground surface
initial velocity of object = 0 m/s
The final velocity of object is v
Work done by gravitational force = m g . h
The final kinetic energy = 1/2 m v²
So
Work done by all forces = Change in kinetic energy
m g h = 1/2 m v² - 0
v² = 2 g h