A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10 s. Coefficient of kinetic friction is (given, g =10 ms -2 ):

Question

A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10 s. Coefficient of kinetic friction is (given, g =10 ms -2 ): (A) 0 .50 (B) 0.60 (C) 0.75 (D) 0.25

Tardigrade Admin · Accepted Answer

S=u t+(1/2) a t2 50=0+(1/2) × a × 100 a=1 m / s 2 F-μ m g=m a 30-μ × 50=5 × 1 50 μ=25 μ=(1/2)

$0.50$

$0.60$

$0.75$

$0.25$

$S = u t + \frac{1}{2} a t^{2}$
$50 = 0 + \frac{1}{2} \times a \times 100$
$a = 1 m / s^{2}$
$F - μ m g = ma$
$30 - μ \times 50 = 5 \times 1$
$50 μ = 25$
$μ = \frac{1}{2}$

Q. A block of mass 5kg is placed at rest on a table of rough surface. Now, if a force of 30N is applied in the direction parallel to surface of the table, the block slides through a distance of 50m in an interval of time 10s. Coefficient of kinetic friction is (given, g=10ms−2 ):

0.50

0.60

0.75

0.25

S=ut+21​at2 50=0+21​×a×100 a=1m/s2 F−μmg=ma 30−μ×50=5×1 50μ=25 μ=21​

$0.50$

$0.60$

$0.75$

$0.25$

$S = u t + \frac{1}{2} a t^{2}$
$50 = 0 + \frac{1}{2} \times a \times 100$
$a = 1 m / s^{2}$
$F - μ m g = ma$
$30 - μ \times 50 = 5 \times 1$
$50 μ = 25$
$μ = \frac{1}{2}$