Question

*b) A block is placed on a wet slippery floor. The mass of block is 15 kg. When it is pulled through a string and spring balance, it shows force equal to 3 N. Find the coefficient of friction. (Fs = u mg) ​

Answer 1

Answer:

Solution

verified

Verified by Toppr

Correct options are A) and C)

Let , I=∫sec

2

xcsc

4

xdx

Using integration by parts ,

I=csc

4

x∫sec

2

xdx−∫

dx

d

(csc

4

x)∫sec

2

xdx

=csc

4

xtanx−∫(−4csc

4

xcotx)tanxdx

=csc

4

xtanx+4∫csc

4

xdx

=csc

4

xtanx+4∫csc

2

xcsc

2

xdx

applying integration by parts again.

=csc

4

xtanx+4[csc

2

x∫csc

2

xdx−∫(−2csc

2

xcotx)(−cotx)dx]

=csc

4

xtanx+4[csc

2

x(−cotx)−2∫csc

2

xcot

2

xdx]

=csc

4

xtanx−4csc

2

xcotx+

3

8

cot

3

x

Substitute , csc

2

x=1+cot

2

x and simplify

Then we get ,

I=−

3

1

cot

3

x+tanx−2cotx+C

Therefore , K=−

3

1

L=1 M=−2

Explanation:

MARK me as brainlist