×

#### 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 block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic frictional coefficient is 0.6, what is the frictional force on the block
A block of mass 2kg rests on the inclined plane making an angel of 30 degree with the horizontal. If the kinetic  frictional coefficient is 0.6, what is the frictional force on the block

```
3 years ago

Arun
25768 Points
```							Given, m = 2 kgθ = 300μ = 0.6R = m g cos ​θTherefore, force of friction, F = ​μ R = ​μ m g cos ​θ = 0.6 X 2 X 9.8 X cos 300 = 10.16 N RegardsArun (askIITians forum expert)
```
3 years 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