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A vehicle of mass $1000 kg$ moving with a velocity of $20 m / s$ is brought to rest in $5 \sec$ after covering a distance of $50 m .$ The unbalanced force acting on the vehicle is ____ $N .$

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Detailed Solution

A vehicle of mass

1000 kg

moving with a velocity of

20 m / s

is brought to rest in

5 \sec

after covering a distance of

50 m .

The unbalanced force acting on the vehicle is

4000 N .

Given that,
Mass of the vehicle

(m) = 1000 kg

Initial velocity

(u) = 20 m / s

Final velocity

(v) = 0 m / s

because car come to rest
Time

(t) = 5 seconds

Distance travelled

= 50 m

From the relation of second law of motion

F = ma

Acceleration,

a = \frac{v - u}{t}

= > a = \frac{0 - 20}{5}

= > a = - 4 m / s^{2}

Acceleration,

a = - 4 m / s^{2}

On putting the values in equation, we have

= > F = 1000 \times - 4

= > F = - 4000 N

Negative sign shows retarding force.

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