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A loaded tractor-trailer with a total mass of 5000 kg traveling at 3.0 km/h hits a loading dock and comes to a stop in 0.64 s. What is the magnitude of the average force exerted on the truck by the dock
Answer:
The value is [tex]F = 6508 \ N[/tex]
Explanation:
From the question we are told that
The total mass is [tex]M_t = 5000 \ kg[/tex]
The time taken to come to a stop is [tex]t = 0.64 \ s[/tex]
The initial velocity is [tex]u = 3.0 km/h = 0.833 \ m/s[/tex]
Generally the magnitude of the average force exerted on the truck by the dock is mathematically represented as
[tex]F = \frac{mu - mv}{t}[/tex]
Here v is the final velocity which is zero since the truck came to a stop
[tex]F = \frac{5000* 0.833 - 5000 * 0}{0.64 }[/tex]
=> [tex]F = 6508 \ N[/tex]
The magnitude of the average force exerted on the truck by the dock is 6508N and this can be determined by using the kinematics equation.
Given :
A loaded tractor-trailer with a total mass of 5000 kg traveling at 3 Km/hr hits a loading dock and comes to a stop in 0.64 s.
According to the given data, the initial velocity is 3 km/hr that is 0.833 m/sec and the final velocity is zero.
First, determine the acceleration using the kinematic equation.
v = u + at
0 - 0.833 = 0.64a
[tex]\rm a = - \dfrac{0.833}{0.64}=1.3015\;m/sec^2[/tex]
Now, the magnitude of the average force exerted on the truck by the dock is calculated as:
F = ma
where F is the force, m is the mass, and a is the acceleration.
[tex]\rm F = -1.3015\times 5000[/tex]
F = - 6508N
So, the magnitude of the average force is 6508N.
For more information, refer to the link given below:
https://brainly.com/question/2996858
