A small car of mass 833 kg is parked behind a small truck of mass 1767 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 41 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 1.5 m/s 2 . What is the acceleration of the truck

Answers

Answer 1

Answer: the acceleration of the truck a_t is 0.6911 m/s²

Explanation:

Given the data in the question;

There is no external force on the system; net force on the system is 0

Mass of the truck with the woman M = 1767 kg + 41 kg = 1808 kg

mass of the car m = 833 kg

car acceleration a_c = 1.5 m/s²

now let a_t be the acceleration of the truck in opposite direction

Action force on the car = Reaction force on the car

ma_c = Ma_t

a_t = ma_c / M

we substitute

a_t = (833 × 1.5) / 1808

a_t = 1249.5 / 1808

a_t = 0.6911 m/s²    

Therefore, the acceleration of the truck a_t is 0.6911 m/s²


Related Questions

Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone of frequency f, at a speed of 10 m/s directly towards Jason, but his aim is a bit off. As a result, Jason runs forward towards the speaker at a speed of 6 m/s before catching it. Then, the frequency that Jason hears while running can be written as (m/n)f Hz, where m and n are relatively prime positive integers. Compute m n.

Answers

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  [tex]f\times \frac{340+6}{340-10}[/tex]

= [tex]f\times \frac{346}{330}[/tex]

So m = 346 , n = 330 .

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