Question

A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming that the body to be initially at rest find :
A) its velocity when force stops acting on it.
B) The distance covered in 10 s after force starts acting on it
@TuringTest @lgbasallote @Lachlan1996

Answer 1

help!

Answer 2

Help

Answer 3

@ParthKohli help

Answer 4

which one are u solving a or b

Answer 5

actually sorry no

Answer 6

Come on Lachan come on!

Answer 7

hahaha sorry, one sec

Answer 8

a = v= 8m/s

Answer 9

with b is this assuming that the force stops acting after 4 seconds? and that there is no friction stopping it?

Answer 10

k

Answer 11

then

Answer 12

well then, assuming no friction and that acceleration stops after 4 seconds. you can just use the basic s=vt. =8*10 = 80m

Answer 13

agree? disagree?

Answer 14

okay!

Answer 15

BUDDY are u here

Answer 16

onesec getting help with maths, pm me

Answer 17

F/m = a = 4/2 = 2 m/s^2
v = at = 2 x 4 = 8 m/s

The distance is harder. We have 6 seconds after the force stops acting (at 8 m/s), and the first 4 seconds with an average velocity of 4, so that's 6 x 8 + 4 x 4. Sound right?

Answer 18

\[F=DeltaP/DeltaT\]
P is the linear momentun
T is the time
\[DeltaP=m(V-V0)\]

Answer 19

V is the final velocity after 4s
V0=0 because of the particle is initially at rest

Answer 20

\[V=DeltaT F/m\]

Answer 21

This should be a clearer explanation:
\[F=ma==>a=\frac{F}{m}=\frac{4N}{2kg}=2m/s^2\]For answer a:
\[V_{f}=V_{i}+at=0m/s + (2m/s^2)(4s) = 8m/s\]
For answer b you need to break it into two distances. First, you need the distance traveled while the force acting on (while it was accelerating):
\[d=V_it+\frac{1}{2}at^2=(0m/s)(4s)+\frac{1}{2}(2m/s^2)(4s)^2=16m\]
Next you need to calculate the distance while it is moving at a constant velocity of 8m/s (which it does for 6 seconds):
\[d=vt=(8m/s)(6s)=48m\]Therefore, the total distance will be \[16m+48m=64m\]