Question

What force must the brakes and tires apply to a $2800\mathrm{kg}$ truck going $30meters$ per second to bring it to rest in$8.0seconds$.

• A. $12000\mathrm{N}$
• B. $13000\mathrm{N}$
• C. $10500N$
• D. $12500\mathrm{N}$

Answer 1

The correct option is C

$10500N$

Step 1: Given data

Mass of truck, $m=2800\mathrm{kg}$;
The initial velocity of the truck, $u=30{\mathrm{ms}}^{-1}$;
The final velocity of the truck, $v=0{\mathrm{ms}}^{-1}$ and
Time is taken to come to rest, $t=8\mathrm{s}$

Step 2: Calculations.
According to question acceleration of truck for its motion during $8\mathrm{s}$ will be calculated using the equation of motion, $v=u+at$.
$a=\frac{v-u}{t}$
$a=\frac{0-30}{8}$
$a=-3.75{\mathrm{ms}}^{-2}$
Negative acceleration indicates that the agent(break in this case) works against the existing motion of the truck. So, the Force required to bring the truck to rest is given by,
$F=m\times a$

$F=2800\times -3.75$
$F=-10500N$
$F=10500N$ against the motion of the truck.

Hence, option (c) is correct.