A belt is moving with constant speed 10 m/s as shown. If a block of 2 kg is dropped on the belt from above, then the timer after which relative motion between block and belt disappears is (Given coefficient of friction between block and belt is 0.2).

t = 2s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t = 3s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t = 4s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t = 5s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D t = 5s
Due to friction on rightward force will act on the block which will accelerate the block when it reaches to velocity V. The relative sleeping will stop ocuring.
$a = μ g a = 2 m / s^{2} v = 10 m / s v = u + a t 10 = 0 + 2 t t = 5 s e c$

Suggest Corrections

Similar questions

Q. A block is gently placed on a conveyor belt moving horizontally with constant speed. After

t = 4 s

, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction between the block and the belt is

μ = 0.2

, then the velocity of the conveyor belt is

Q. A block is gently placed on a conveyor belt moving horizontally at a constant speed. After

t = 4 s

, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of static friction between the block and the belt is

μ = 0.2

, then what is the velocity of the conveyor belt ?

Q. A belt is moving with constant speed u (clockwise, as shown in figure) on two fixed wheels. The length of the belt between

ℓ = 10 m

, as shown. At time t = 0, a small block starts moving from the upper right end of the belt with leftward speed v (w. r. t. ground).

μ

is friction coefficient between belt and block. The moving part of the belt between the wheels is always horizontal.

Q. Two 30 kg blocks rest on a massless belt which passes over a fixed pulley and is attached to a 40 kg block. If coefficient of friction between the belt and the table as well as between the belt and the blocks B & block C is

μ

. The system is released from rest from the position shown, the speed with which the block B falls off the belt is

An object is gently placed on a long conveyer belt moving with a uniform speed of $5 m s^{- 1}$ . If the coefficient of friction between the block and the belt is 0.5, how far will the box move on the belt (relative to belt) before coming to rest? Take $g = 10 m s^{- 2}$ .

Join BYJU'S Learning Program

A belt is moving with constant speed 10 m/s as shown. If a block of 2 kg is dropped on the belt from above, then the timer after which relative motion between block and belt disappears is (Given coefficient of friction between block and belt is 0.2).

The correct option is D t = 5sDue to friction on rightward force will act on the block which will accelerate the block when it reaches to velocity V. The relative sleeping will stop ocuring.a=μga=2m/s2v=10m/sv=u+at10=0+2tt=5sec

The correct option is D t = 5s
Due to friction on rightward force will act on the block which will accelerate the block when it reaches to velocity V. The relative sleeping will stop ocuring.
$a = μ g a = 2 m / s^{2} v = 10 m / s v = u + a t 10 = 0 + 2 t t = 5 s e c$