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Viefleur [7K]
3 years ago
13

According to the third law of motion when we put an object, the object pushes back on us with an equal and opposite force. If th

e object is massive truck park along the roadside, it will probably not move. A student justify this my answering that the two opposite and equal forces cancel Each Other. Comment on this logic and explain why the truck do not move?
Physics
1 answer:
vichka [17]3 years ago
5 0


Hi pupil Here's Your answer :::


➡➡➡➡➡➡➡➡➡➡➡➡➡


Student's justification is not correct. Two equal and opposite force cancel each other if the act on the same body. According to the third law of motion action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other.

When we push a message track, then the applied force on the truck is not sufficient to overcome the force of friction between the tyres of truck and ground, hence, truck does not move.


⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅



Hope this helps .
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(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.

(c)Velocity at the end of that time will be 8.5 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f  = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f  = 85 \ m

c)

The velocity at the end of that time is found as;

\rm  v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5 m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

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