Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
ΔK = -6 10⁴ J
Explanation:
This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved
initial instant. Before the crash
p₀ = m v₁ + M 0
final instant. Right after the crash
p_f = (m + M) v
p₀ = p_f
mv₁ = (m + M) v
v = [tex]\frac{m}{m+M} \ v_1[/tex]
we substitute
v = [tex]\frac{20}{20+40}[/tex] 3
v = 1.0 m / s
having the initial and final velocities, let's find the kinetic energy
K₀ = ½ m v₁² + 0
K₀ = ½ 20 10³ 3²
K₀ = 9 10⁴ J
K_f = ½ (m + M) v²
K_f = ½ (20 +40) 10³ 1²
K_f = 3 10⁴ J
the change in energy is
ΔK = K_f - K₀
ΔK = (3 - 9) 10⁴
ΔK = -6 10⁴ J
The negative sign indicates that the energy is ranked in another type of energy
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
