A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the lad
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Newton's second law we can find that the correct answer is:
E) It cannot be determiner whick block has more masses from the information provided
Newton's second law establishes the relationship between force, mass, and acceleration of a body. Since force and acceleration are vector quantities, their components must be added on each axis
For this problem we have two bodies, let's write Newton's second law for the body B, we assume that the body B descends
W_b - T = m_b a
W_b = m_b g
m_b - T = m_b a
Where W_b is the weight of block B, T the tension of the string, mb the mass of block b and the acceleration
Now let's find the relation for block A
let's set a datum with the x axis parallel to the ramp
T - Wₓ = mₐ a
sin θ = Wₓ / W
Wₓ = Wₐ sin θ
Wₐ = mₐ g
Where Wₓ is the component of the weight, Wₐ the weight of the body A and θ the angle of the plane
Let's write our system of equations
m_b g - T = m_b a
T - mₐ g sin θ = mₐ a
let's add the equations
g (m_b - mₐ sin θ) = (m_b + mₐ) a
a =
Let's analyze this expression
- The numerator is positive the body B descends, this occurs when
m_b - mₐ sin θ > 0
- The numerator is negative, body B rises
m_b - mₐ sin θ <0
We can observe that the acceleration is positive or negative depending on the relation of the masses and the angle of the plane.
In conclusion using Newton's second law we find that the correct answer is
E ) It cannot be determiner whick block has more masses from the information provided
learn more about Newton's second law here:
Complete Question:
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.
Answer:
a) Speed of car A at the start of sliding = 4.23 m/s
b) speed of car B at the start of sliding = 3.957 m/s
c) Speed of car B before the collision = 7.28 m/s
Explanation:
NB: The figure is not provided but all the parameters needed to solve the question have been given.
Let the frictional force acting on car A, ............(1)
Since frictional force is a type of force, we are safe to say .......(2)
Equating (1) and (2)
a) Speed of A at the start of the sliding
b) Speed of B at the start of the sliding
Let the speed of car B before collision =
Momentum of car B before collision =
Momentum after collision =
Applying the law of conservation of momentum: