IIT JEE , AIEEE Forum - Physics , Mechanics

asish

any one having DCPANDEY MECHANICS VOL 1 pls solve introductary exercise 6.3 problem 5 of 'WORK POWER ENERGY' pls help me out
i cannot draw fig so iam giving the qn no.

uday_zingtudor

hey easy one!!

To overcome frictional force and to move the block up,

mgcos37=Mg

M= 3m/5

if u dint understand i'll xplain

edit: the normal force N acting on small block is mgcos37

frictional force acting downward is

N =

mgcos37 = 3mg/5

The spring has nothing to do!!!

The tension acting upwards is T = Mg

T >

N .For the minimal case, T =

N

that implies 3mg/5=Mg

M= 3m/5

asish

here's the qn without fig
a block of nass 'm' is attached with a massless spring of force const 'k' &placed on a rough inclined plane of inclination 37 degrees &coeff of friction=3/4.find min value of 'M' to move block up the plane.
DESCRIPTION OF FIG:'m' is on incinedplane attached to a spring which is attached to a string which passes over a pulley and other end tied to mass 'M' which hangs
hope u understand

asish

explain

uday_zingtudor

u've asked me 6.3 1st question

the spring extends by x and the block A comes down by x.Let it's mass be M

conservation of energy Mgx=1/2kx^2

kx=2Mg

the block B of mass m is hold just above the ground

force acting upward is kx and downward is mg

kx=mg

so m=2M

M=m/2

~~Cheerio!!!

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