Question

- A stone of 1 kg is thrown with a velocity of 20 m s' across
the frozen surface of a lake and comes to rest after travelling
a distance of 50 m. What is the force of friction between the
stone and the ice?​

Answer 1

Explanation:

Initial velocity of the stone, u= 20 m/s

Final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

We know the third equation of motion

v²=u²+2as

Substituting the known values in the above equation we get,

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

We know that

F = m×a

Substituting value of a= -4 in F = m x a

we get,

F = 1×(-4) = -4N

(Here the negative sign indicates the opposing force which is Friction)

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Answer 2

Given :

A stone of 1 kg is thrown with a velocity of 20 m/s across  the frozen surface of a lake and comes to rest after travelling  a distance of 50 m

To Find :

Force of friction

Solution :

$\sf \bigstar\ \; \pink{v^2-u^2=2as}$

where ,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes displacement / distance

$\sf \bigstar\ \; \purple{F=ma}$

where ,

• F denotes force
• m denotes mass
• a denotes acceleration

________________________

Mass , m = 1 kg

Initial velocity , u = 20 m/s

Final velocity , v = 0 m/s

∵ Comes to rest .

Distance , s = 50 m

$\sf v^2-u^2=2as\\\\\to \sf (0)^2-(20)^2=2a(50)\\\\\to \sf 0-400=100a\\\\\to \sf -400=100a\\\\\leadsto \sf a=-4\ m/s^2\ \; \bigstar$

Note : Negative sign of acceleration denotes retardation / deceleration

$\sf F=ma\\\\\to \sf F=(1)(-4)\\\\\leadsto \sf F=-4\ N\ \; \bigstar$

Note : Negative sign of force denotes retarding force , which is nothing but frictional force .

Net Force = Frictional force = - 4 N