Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
372 views
in Physics by (84.6k points)
closed by
A solid block of mass `m = 1 kg` is reasting on a horizontal platform as shown in figure. The `z` direction is vertically up. Coefficient of friction between the block and the paltform is `mu = 0.2`. The platform is moved with a time dependent velocity given `vec(V) = (2that(i) + that(j) + 3hatk) m//s`. Then the magnitude of the net force exerted by the block on the platform is : (Take `g = 10 m//s^(2)`)
image
A. `sqrt(168)N`
B. `sqrt(174)N`
C. `sqrt(194)N`
D. None of these

1 Answer

0 votes
by (87.2k points)
selected by
 
Best answer
Correct Answer - B
Acceleration of the platform
`overset(vec)(a_(p) ) - (doverset(vec)v)/(dt) - 2hati + hat(j) + 3hat(k)`
Horizontal force on the block
Normal force on the block
`a_(H) = sqrt(4 + 1) = sqrt(5) m//s^(2)`
`a_(v)= 3 m//s^(2)`
Normal force on the block
`N = m (g + a_(v)) = 1 xx 13 = 13N`
Maximum acceleration that friction can positive
`a_(mass) = mu(g + a_(y)) = 0.2 xx 13 = 206 m//s^(2)`
`:. a_(max) gt a_(H)`
`:.` Value of frication force on the block
`f = ma_(H) = sqrt(5)N`
`:.` Force by the platform on the block is
`F = sqrt(N^(2) t^(2)) = sqrt(169 + 5) = sqrt(174)N`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...