Question

2 blocks of equal weight are connected by a string. If block slides with constant velocity, find the force of kinetic friction.

Answer 1

Answer:

Given that,

Mass of block A=10 kg

Kinetic friction μ=0.2

Now, according to diagram

Now, the component of the weight in the direction of the incline

W=m

A

gsinθ

T=F

f

−W....(I)

When the force of friction

F /f

=μ

k N

F/f

=μ k

(m /A

gcosθ)

Now, put the value in equation (I)

The tension is

T=mA

gsin30 0

−μ /k

m A

gcos30 0

We know that,

T−m

B

g=0

T=m B g0

So, m B

g=m A

g(sin30 0

−0.2×cos30 /0 ) m B

=10×( 2/1 −0.2×2/3 ) m B

=10×(0.5−0.1732) m B

=10×0.3268 m B

=3.268 m B

=3.3kg

Hence, the mass of block B is 3.3 kg