A block of mass 1kg is given a velocity of 5m/s on a horizontal flat surface. What will be the change in internal energy of the block-surface system, when the block comes to rest. The coefficient of kinetic friction is 0.3 and coefficient of static friction is 0.4.a)25Jb)12.5Jc)0d)30JCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Given:
Mass of the block, m = 1 kg
Initial velocity of the block, u = 5 m/s
Coefficient of kinetic friction, μk = 0.3
Coefficient of static friction, μs = 0.4

To find:
Change in internal energy of the block-surface system when the block comes to rest

Formula:
The work done by friction can be calculated using the formula:
Work done by friction, W = force of friction × distance

The force of friction can be calculated using the formula:
Force of friction, f = coefficient of friction × normal force

The normal force can be calculated using the formula:
Normal force, N = mass × acceleration due to gravity

The distance can be calculated using the formula:
Distance, s = (final velocity)^2 - (initial velocity)^2 / 2 × acceleration

Calculation:
1. Calculation of normal force:
N = m × g
= 1 kg × 9.8 m/s²
= 9.8 N

2. Calculation of distance:
The block comes to rest, so the final velocity, v = 0 m/s
s = (v^2 - u^2) / (2a)
= (0^2 - 5^2) / (2 × a)
= -25 / (2 × a)
= -12.5 / a

3. Calculation of acceleration:
The block is coming to rest, so its acceleration, a = -μk × g
= -0.3 × 9.8
= -2.94 m/s²

4. Calculation of distance:
s = -12.5 / (-2.94)
≈ 4.25 m

5. Calculation of force of friction:
f = μk × N
= 0.3 × 9.8
= 2.94 N

6. Calculation of work done by friction:
W = f × s
= 2.94 × 4.25
≈ 12.5 J

Therefore, the change in internal energy of the block-surface system when the block comes to rest is 12.5 J. Hence, the correct answer is option B.

Answer

The kinetic energy of the block will change into internal energy of the system.
Initial kinetic energy = 0.5*1*25 = 12.5J.
Therefore the internal energy change = 12.5J.

Similar NEET Doubts

On the horizontal surface of a truck, a block of mass 2 kg is placed and the coefficient of friction between the block and surface is (μs = 0.6) truck starts moving with an acceleration 4 m/s2 , the frictional force on the block will be? (g = 10 m/s2)

A body of mass 2kg is drawn with a velocity of 2m/s on a horizontal table .if the coefficient of kinetic friction is 0.3 then calculate the heat produced in 5 sec. g=9.8 m /s^2 1)5. 2)10. 3)14. 4)28?

On the horizontal surface of truck a block of mass 2 kg is placed and coefficient of friction between the block and surface is . If truck starts moving with acceleration 4 m/s² then frictional force on the block will be?

A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest?

A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then the maximum and minimum value of force f for which the block remains at rest are?

Top Courses for NEETView all

Biology Class 11

Daily Test for NEET Preparation

NEET Mock Test Series 2025

Topic-wise MCQ Tests for NEET

Physics Class 11

Top Courses for NEET

Top Courses for NEET

Suggested Free Tests

Related Content

About NEET

NEET Preparation Material

How to Prepare for NEET Exam?

NCERT Based Study Material

NEET Mock Tests

Other Popular Exams

Company