# Thread: science- force mass and acceleration

1. ## science- force mass and acceleration

Hey guys please work this out for me showing me the steps and explaining it to me- science is so not my forte.

g= 9.8m/s/s => this is what it says on top of the sheet. Do I really need it?
Neglect friction. FIND THE FORCE REQUIRED.

1) For a 50kg woman on a 20kg bicycle travelling at 6m/s to come to a halt in 2 seconds.

2) To give a mass of 25 kgs. an acceleartion of 18m/s/s.

3) To accelerate a 1500 kg space probe from 54 000 to 54900 kms/hr in 20 seconds.

4) To increase the velocity of a 1 00 kg car from 10m/s to 20m/s in 5 seconds.

2. Originally Posted by delicate_tears
Hey guys please work this out for me showing me the steps and explaining it to me- science is so not my forte.

g= 9.8m/s/s => this is what it says on top of the sheet. Do I really need it?
Neglect friction. FIND THE FORCE REQUIRED.

1) For a 50kg woman on a 20kg bicycle travelling at 6m/s to come to a halt in 2 seconds.

2) To give a mass of 25 kgs. an acceleartion of 18m/s/s.

3) To accelerate a 1500 kg space probe from 54 000 to 54900 kms/hr in 20 seconds.

4) To increase the velocity of a 1 00 kg car from 10m/s to 20m/s in 5 seconds.
$\displaystyle F=ma$

$\displaystyle F$ is the force required or net force.
$\displaystyle m$ is the total mass.
$\displaystyle a$ is the rate of change in velocity.

1)

$\displaystyle F=ma$
$\displaystyle F=m\left(\frac{v-u}{t}\right)$
$\displaystyle F=(50+70)\left[\frac{(0)-(6)}{(2)}\right]$
$\displaystyle F=(120)(-3)$
$\displaystyle F=-360N$

2)

$\displaystyle F=ma$
$\displaystyle F=(25)(18)$
$\displaystyle F=450N$

3)

$\displaystyle \Delta v$
$\displaystyle =54900-54000$
$\displaystyle =900 kmh^{-1}$
$\displaystyle =\left(\frac{900\times 1000}{1\times 60\times 60}\right)ms^{-1}$
$\displaystyle =250ms^{-1}$

$\displaystyle F=ma$
$\displaystyle F=m\left(\frac{\Delta v}{t}\right)$
$\displaystyle F=(1500)\left(\frac{250}{20}\right)$
$\displaystyle F=18750N$

4)

$\displaystyle F=ma$
$\displaystyle F=m\left(\frac{v-u}{t}\right)$
$\displaystyle F=(100)\left[\frac{(20)-(10)}{5}\right]$
$\displaystyle F=200N$