Click to Chat

1800-2000-838

+91-120-4616500

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
        A particle moves along the X-axis according to the law x=at2-bt3,where a and b are positive constants.At moment t=0,force acting on the particle is equal to F0.Find the value of force Fx at points of turn and at moment when particle again turn up at point x=0.
one year ago

Khimraj
2612 Points
							Let mass of particle is m.x = at2 – bt3particle will be at x= 0 at time0 =t2(a-bt)t = 0, a/b secvelocity of paerticlevx= dv/dx = 2at -3bt2particle will turn when velocity of particle will be zero at a point another then starting point0 = t(2a-3bt)t = 2a/3baccelaration of particleax= dvx/dt = 2a – 6btFx when particle turn = m*(ax at t = 2a/3b) = m*(2a – 4a) = –2ma $\hat{i}$Fx when particle again turn up at initial point = m*(ax at t = 2a/3b) = m*(2a – 6a) = -4ma$\hat{i}$Hope it clears.

one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions