Friction Numerical on Coefficient Class 11 Nootan ISC Physics Solutions

Friction Numerical on Coefficient Class 11 Nootan ISC Physics Solutions Ch-7.  Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website for detail information about ISC Board Class-11 Physics.

Numericlas Based on Friction, Coefficient of Friction and Motion on a Rough Level Surface

Ch-7 Friction Class 11 ISC Nootan Solutions of Kumar and Mittal Physics of Nageen Prakashan

 Board ISC Class 11 Subject Physics Writer Kumar and Mittal Publication Nageen Prakashan Chapter-7 Friction Topics Numericals on Friction, Coefficient of Friction and Motion on a Rough Level Surface Academic Session 2024-2025

Friction Numerical on Coefficient Class 11 Nootan ISC Physics Solutions

Question-1: A body rolled on a surface with a speed of 10 m/s comes to rest after covering a distance of 50 m. Find the coefficient of friction. Take g = 10 m/s².

s = 50 m    ,    u = 10 m/s

Now, using the third equation of motion, we get:

=> v^2-u^2=2as

Now, putting the values, we get:

=> 0^2 – 10^2 = 2 * a * 50

-100 = 100 a

a = -1

The negative sign of a shows it is retardation.

Now, F = ma

F = m * 1 = m

So, the frictional force is m kg m s^(-2)

We know, the formula of force of friction, F= k m g

where k = coefficient of friction, m = mass, and g= acceleration due to gravity.

Now, F=k m g

Putting the calculated value of F, we get:

m=k m * 10

k = 0.1

Question-2: A block of mass 1.5 kg placed on a rough horizontal surface is pulled by a constant horizontal force of 1.2 kg-f. The coefficient of friction between the block and the surface is 0.3. Find the acceleration produced in terms of g.

Answer- N = 1.5 g    ,     f = 0.3 x 1.5 x g = 0.45 g

∴ Net force = 1.2 g – 0.45 g = 0.75 g

0.75 g = 1.5 x a

a = 0.5 g = g /2.

Question-3: Two blocks, each of mass 3.0 kg, are connected by a light cord and placed on a rough horizontal surface. When a horizontal force of 20 N is applied on a block, an acceleration of 0.50 m/s² is produced in each block. Find the tension in the string and the frictional force which is same on the two blocks.

Answer- Total external force = 6 x 0.5 = 3 N

Net force = 20 – 2f = 3 => – 2f = 3 – 20 => 2f = 17

=> f = 8.5 N

Now for block

=> T – f = 3 x 0.5 = 1.5

=> T = 1.5 + f

Now putting the value of f.

=> T = 1.5 + 8.5 = 10 N

Question-4: A car is moving on a straight road at a speed of 20 m/s. The coefficient of friction between the tyres of the car and the road is 0.4. Find the shortest distance within which the car can be stopped. (g = 10 m/s²)

Answer- initial speed U = 20 m / s    ,    Final velocity F = 0

We take g = 10 m/s²

We need to find the acceleration which will be negative given that the car is coming to rest.

A = – 0.4 × 10 = -4 m / s²

From the formula :

V² = U² + 2as

We can get S

0 = 20² – 2× 4 × S

0 = 400 – 8S

S = 400/8 = 50m

Question-5: A car going at a speed of 7 m/s can be stopped by applying brakes in a shortest distance of 10 m. Show that the total frictional force opposing the motion, when brakes are applied, is 1/4th of the weight of the car. (g = 9.8 m/s²)

u = 7 m/s; v = 0; s = 10 m.

Using, v² = u² + 2as, we get

a=(v² -u²)/2s = o-49/20 = -2.45m/s²

If we consider, g = 9.8 m/s², then we get

a=-g/4

Thus, the resistance to the motion, that is resistive force is

-F=-ma=mg/4

This is one fourth the weight of the car.

Question-6: A fireman of mass 80 kg slides down a pole during a rescue mission. The force of friction is constant at 720 N. Find the acceleration of the man. (g = 10 m/s²)

Answer- Force of friction is upward = 720 N

Force of gravity downward = M*g = 80*10 = 800 N

So resultant force = 800 -720 = 80 N

F = m * a

80 = 80* a

a = 1.0 m/s²

—:  end of Friction Numerical on Coefficient Class 11 Nootan ISC Physics Solutions Ch-7 :—

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