Question

A constant force of friction 50N is acting on a body of mass 200 kg moving initially with a speed of 15m/s. How long does the body take to stop ? What distance will it cover before coming to rest?

Answer 1

F=-50N
M=200kg
U=15m/s
F=Ma
a=F/M=-50/200=-0.25
Now using
V^2-U^2=2aS
0-(15)^2=2(-0.25)S
S=-225/-0.5=450m
As V=U+at
0=15-0.25t
t=-15/-0.25=60s

Answer 2

Given :

Force (F) = 50 N

Mass (m) = 200 Kg

Initial speed (u) = 15 m/sec

To Find :

(i) How long does the body take to stop ?

(ii) What distance will it cover before coming to rest?

Solution :

We know,

    Force (F) = Mass (m) × Acceleration (a)

∴             a  =  $\frac{50}{200}$ = 0.25 m/sec²

(i) Using Kinematic equation for uniformly accelerating body, we have

   Final velocity (v)  = Initial velocity (u) - Acceleration (a) × Time (t)

⇒     0                      =   15 - 0.25t

⇒    0.25t                =    15

⇒            t                =    $\frac{15}{0.25}$

∴             t                = 60 sec

Therefore, the time taken by the body to stop is 60 seconds.

(ii) Using Kinematic equation for uniformly accelerating body, we have

    v²   =  u²  - 2as                       (s = distance)

⇒  0       =  (15)²  - 2×0.25×s

⇒ 0.5s   =   225

∴        s   =  450 m

Therefore, the distance travelled by the body before coming to frest is 450 m.