Question #c843a

Question

1325 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-10T15:59:38

$v=25ms^-1$

$m=1500kg$
$v_(0)=15ms^-1$
$F_1=1000N$
$t=15s$

Before a force acting on it, the object is at constant velocity (assuming no other forces is acting on it: $eg.$ friction), thus
$NetF=0$

After a force acting on it, the object experiences a net force of same magnitude as the applied force.
$F=ma$
$1000=1500a$
$a=2/3 ms^-2$

$v=v_0+at$
$v=15+2/3*15$
$v=15+10$
$v=25ms^-1$

Answer 2 · 2017-07-10T16:10:41

Its velocity will be 25m/s

We can use two known equations to solve this problem.
These are
$F=m*a$
$v=u +a*t$

Lets substitute the values
F=1000N
m=1500 kg
then we have $a=F/m=1000/1500$ =2/3 m/ $s^2$

Also
u=15 m/s
t=15 s
so that $v=u+a*t=15 + 2/3*15=15+10=25s$