A block of mass 25 kg is pulled along a horizontal surface by aforce at an angle 45° with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The work done by the applied force durin

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A block of mass 25 kg is pulled along a horizontal surface by a force at an angle 45° with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The work done by the applied force during a displacement of 5 m of the block is: [2025]

1 970 J

2 735 J

3 245 J

4 490 J

Ans.
(3)

$N = m g - \frac{F}{\sqrt{2}}$

Block travels with uniform velocity

$\frac{F}{\sqrt{2}} = μ [mg - \frac{F}{\sqrt{2}}]$

$\frac{F}{\sqrt{2}} = 0.25 [25 \times 9.8 - \frac{F}{\sqrt{2}}]$

$\Rightarrow 1.25 \frac{F}{\sqrt{2}} = 61.25$

$\Rightarrow F = \frac{61.25 \times \sqrt{2}}{1.25} = 49 \sqrt{2} N$

$W_{ext} = F S \cos 45 ° = 49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}} = 245 J$ .

Want Us to Email you About Special Offers & Updates?