#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Plz help me solve 4,5 plz tell me ho w to approach these problems

Chandan
121 Points
3 years ago
TRICKS AT THE END:
for 4:
to move block A towards right force F+umg should be slightly greater than spring force for convineance we will take it equal......................(1)
if the system is released from rest block B will move down
let us assume it moves down by x meters.
so will the spring be extended by x meters.
from statement 1:
kx=F+umg …............(2)    (k is the spring constant assumed,will cancel afterwards)
and as block goes doun by x meters,it will loose Mgx joules of energy which will be stored in the spring.
so kx2\2=Mgx
kx=2Mg
putting the value of kx in eq 2
2Mg=F+umg
M=F\2g+um\2
TRICK:
NOTE:
not applicable for all but for many.
identify the forces on each component and write them down now make all the components in one line like make the block B horizontal at the level of others
now as the forces are written on them,simple tension and f=ma will solve the problem.

for:5
the distance travelled =360\60 *r
=4pi\9
take the kinetic energy at that point
and equate it to mv2\2 find v
centripetal accelaration=v2\r
using v2=u2+2as
find a
and total accelaration = centripetal acceleration+a;
and hence find the angle using vestor method
UNFORTUNATELLY NO TRICk FOR SUCH TYPE OF QUESTIONS
hope it helped