5.35 A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then moves with uniform velocity.

Discuss the motion of the block as viewed by

(a) A stationary observer on the ground

(b) An observer moving with the trolley

0 11 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    Mass of the block = 15 kg

    Coefficient of static friction between the block and the trolley  μ = 0.18

    Acceleration of the trolley = 0.5 m/s2

    (a) Force experienced by block, F = MA = 15 × 0.5 = 7.5 N. This fore acts in the direction of motion of the trolley

    Force of friction, Ff = μmg = 0.18 ×15×10 N = 27 N

    Force experienced by the block is less than the friction, hence for a stationary observer on the ground, the block will be stationary

    (b) When an observer moves with the trolley, the trolley will appear to be at rest

Similar Questions for you

A
alok kumar singh

Kindly go throughb the solution

 

A
alok kumar singh

Kindly go through the solution

 

A
alok kumar singh

 

= 0.92 * 1260 = 1161 m/s

A
alok kumar singh

For 2 kg block

T – 2g sin37 = 2a        . (i)

For 4 kg block

4g – 2T =  4 a 2

2g – T = a                    . (ii)

T = (2g – a)

2g – a – 2g ×  3 5  = 2a

3a = 2g ×  2 5

a = 4 g 1 5

A
alok kumar singh

a = ( m 1 m 2 ) g ( m 1 + m 2 ) = g 8

8m1 – 8m2 = m1 + m2

7m1 = 9m2

m 1 m 2 = 9 7          

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post