Question

An iron sphere of mass 1 kg is moving with a velocity of 20m/sec on a cemented floor. It comes to rest after travelling a distance of 50m. Find the force of friction between the sphere and the floor.

Answer 1

let initial velocity be u
     final velocity be v
     acceleration be a
     mass= 1000g.
     distance= 50m
   acc. to formula v2=u2-2a*s
     so,   0 = 400- 2*a*50
             -400= -100*a
              a = 4 m/s2
          F = MASS * ACCELERATION
           F =  1000*4
               = 4000N.

Answer 2

mass of sphere, m = 1kg
initial velocity, u = 20 m/s
final velocity, v = 0
distance travelled, s = 50m
let the friction acting = f 

Here, the kinetic energy of the ball is being dissipated to overcome friction. So, according to conservation of energy,

the change in kinetic energy = work done by friction

$\frac{1}{2}m(v^2-u^2)=f.s\\ \\ \Rightarrow \frac{1}{2} \times 1 \times (0^2-20^2)=f \times 50\\ \\ \Rightarrow \frac{1}{2} \times (-400)=50f\\ \\ \Rightarrow 50f=-200\\ \\ \Rightarrow f = - \frac{200}{50} \\ \\ \Rightarrow f = -4\ N$

So friction acting is 4N. The negative sign denotes that it acts in the direction opposite to the motion of the sphere.

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using equations of motion,
let the acceleration due to friction = a
v² - u² = 2as
⇒ 0 - 20² = 2×a×50
⇒ -400 = 100a
⇒ a = -400/100
⇒ a = -4 m/s²

mass of sphere = 1 kg
friction = ma = 1×4 = 4N