#### Thank you for registering.

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

Click to Chat

1800-1023-196

+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 mass of M kg is suspended to a weightless string.Find the horizontal force that is required to displace it until it makes an an angle 45 degree with initial vertical direction.

Ashish Kumar
askIITians Faculty 32 Points
7 years ago
The work done by the force will be equal to the potentail energy gained by the mass.
Let L be the length of the string ,
F.d= MgL (1 -cos45 )
The horizontal displacement of the mass is d= L sin 45
F L sin45 = Mg L(1 - 45)
F = Mg (sq. root 2 - 1)

Lokesh
24 Points
7 years ago
USE CONSERVATION OF ENERGY
shraman
19 Points
3 years ago
Let x be the force required
T Sin 45=M kg   …....1
T Cos45=x    ….......2
Divide equation 1 and 2,
1=M/x
x=M
askIITians Faculty 629 Points
9 months ago
Dear student,
Please find the solution to your problem attached herewith.

The work done by the force will be equal to the potentail energy gained by the mass.
Let L be the length of the string ,
F.d= MgL(1 – cos45)
The horizontal displacement of the mass is d = Lsin45
FLsin45 = MgL(1 – cos45)
F = (2 - 1)Mg

Hope it helps.
Thanks and regards,
Kushagra