Answer:
(a) the coefficient of friction is 0.451
This was derived using the energy conservation principle (the total energy in a closed system remains constant).
(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.
Explanation:
For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.
Thank you for your attention; I trust this is beneficial to you.
Answer:
B. Truck X was ahead, not truck Y.
Explanation:
Let's analyze the information provided.
Truck X moved from the point (0,20) to (2.8,50). This indicates that it began at the 20th kilometer and reached 50 km in 2.8 hours. Thus, its speed is v1 = (s2 - s1) / t
v1 = (50 - 20) / 2.8
v1 = 10.7 km/h
Given that it started from the 20th km, it indeed had a head start. Since the line on the graph is linear, this shows its speed was constant without any change in direction.
On the other hand, Truck Y's movement went from the origin (0,0) to (5,20), meaning it took 5 hours to travel 20 km, resulting in a speed of v2 = 20 / 5
v2 = 4 km/h
Again, the straightness of its graph line signifies it maintained a constant speed in a single direction.
Thus, it is evident that Rosa erred in her assumption that Truck Y had a head start.
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
Answer:
What is your question?
Please feel free to modify or inquire in the comments.
Thank you.
Explanation:
Answer:
Speeds of 1.83 m/s and 6.83 m/s
Explanation:
Based on the law of conservation of momentum,
where m represents mass, 
is the initial speed before impact, 
and 
are the velocities of the impacted object after the collision and of the originally stationary object after the impact.
Thus, 
After the collision, the kinetic energy doubles, therefore:
Substituting the initial velocity of 5 m/s provides the equation needed to proceed.
We know that leads to 
Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25. 
By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.
