A 1.2-kg object moving with a speed of 8.0 m/s collidesperpendicularly with a wall and emerges with a speed of 6.0 m/s inthe opposite direction. If the object is in contact with the wallfor 2.0 ms, what is the magnitude of the average force on theobject by the wall?
a. 9.8 kNb. 8.4 kNc. 7.7 kNd. 9.1 kNe. 1.2 kN

Respuesta :

Answer:

8.4 kN

Explanation:

Change in momentum = m₁v₁- m₂v₂ = m(v₁-v₂) = 1.2[8 - (-6)] = 16.8kgm/s

v₂ has been taken as negative because velocity of object colliding ball as +ve. Both are in opposite directions.

Change in momentum = impulse = Force × Time

Time = 2 ms = 2 milli seconds = 0.002 s.

Force = Change in momentum/Time = 16.8/0.002 = 8.4kN

The appropriate answer is Option b (8.4 kN).

Given values are:

Mass of the object,

m = 1.2 kg

Initial speed,

u = 8 m/s

Final speed,

v = -6 m/s

Time,

t = 2 ms

or,

 = [tex]2\times 10^3 \ s[/tex]

Now,

→ The Force is:

[tex]Force = \frac{Impulse}{Time}[/tex]

          [tex]=\frac{m(v-u)}{t}[/tex]

By substituting the values, we get

          [tex]=\frac{1.2(8+6)}{2\times 10^{-3}}[/tex]

          [tex]=\frac{1.2\times 14}{2\times 10^{-3}}[/tex]

          [tex]=8400 \ N[/tex]

or,

             [tex]=8.4 \ kN[/tex]

Thus the above is the correct solution.

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