A vehicle travelling at $30 m / s$ stops at a distance of $50 m$ decelerating uniformly. A force of $9000 N$ is exerted on it by the brakes. What is the work done by the brakes?

4.5 \times 10^{4} J

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2.7 \times 10^{5} J

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4.5 \times 10^{5} J

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2.7 \times 10^{4} J

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Solution

The correct option is D $4.5 \times 10^{5} J$
The work done by the brakes is calculated using the relation ,
$W = F \times s$ ,
where $F$ is the force applied and $s$ is the displacement in the direction of force.

Hence, $W = 9000 \times 50 = 4.5 \times 10^{5} N$

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A vehicle travelling at 30 m/s stops at a distance of 50 m decelerating uniformly. A force of 9000 N is exerted on it by the brakes. What is the work done by the brakes?

The correct option is D 4.5×105 J The work done by the brakes is calculated using the relation , W=F×s,where F is the force applied and s is the displacement in the direction of force. Hence, W=9000×50=4.5×105 N

A vehicle travelling at $30 m / s$ stops at a distance of $50 m$ decelerating uniformly. A force of $9000 N$ is exerted on it by the brakes. What is the work done by the brakes?

The correct option is D $4.5 \times 10^{5} J$
The work done by the brakes is calculated using the relation ,
$W = F \times s$ ,
where $F$ is the force applied and $s$ is the displacement in the direction of force.

Hence, $W = 9000 \times 50 = 4.5 \times 10^{5} N$