Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
Momentum is conserved in a collision. Momentum is mass*velocity, so you can find your answer by calculating initial and final momentums and setting them equal to each other.
15kg * 3.50 m/s + 9kg * 2.35 m/s = 73.65 kg m/s
73.65 = 9 * 2.8 + 15x
solve for x
x= 3.23
The final velocity is 3.23 m/s
Answer:
Explanation:
(a) Work done, W = 1.82 x 10^4 J
(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)
(c) According to the first law of thermodynamics
Q = W + U
Q = 1.82 x 10^4 - 4.07 x 10^4
Q = - 2.25 x 10^4 J
