If a block moving up an inclined plane at 30° with a velocity of 5 m / s, stops after 0.5 s, then coefficient of friction will be nearly

Question

If a block moving up an inclined plane at 30° with a velocity of 5 m / s, stops after 0.5 s, then coefficient of friction will be nearly (A) 0.5 (B) 0.6 (C) 0.9 (D) 1.1

Tardigrade Admin · Accepted Answer

Using v=u+a t retardation will be provided by friction as well as gravitational force a=(u/t) g sin 30°+μ g cos 30°=(5/0.5)=10 μ=(1/√3) ≅ 0.6

Q. If a block moving up an inclined plane at $3 0^{\circ}$ with a velocity of $5 m / s$ , stops after $0.5 s$ , then coefficient of friction will be nearly

0.5

0.6

0.9

1.1

Using $v = u + a t$
retardation will be provided by friction as well as gravitational force
$a = \frac{u}{t}$
$g sin 3 0^{\circ} + μg cos 3 0^{\circ} = \frac{5}{0.5} = 10$
$μ = \frac{1}{3} ≅ 0.6$

Q. If a block moving up an inclined plane at 30∘ with a velocity of 5m/s, stops after 0.5s, then coefficient of friction will be nearly

0.5

0.6

0.9

1.1

Using v=u+at retardation will be provided by friction as well as gravitational force a=tu​ gsin30∘+μgcos30∘=0.55​=10 μ=3​1​≅0.6

Q. If a block moving up an inclined plane at $3 0^{\circ}$ with a velocity of $5 m / s$ , stops after $0.5 s$ , then coefficient of friction will be nearly

Using $v = u + a t$
retardation will be provided by friction as well as gravitational force
$a = \frac{u}{t}$
$g sin 3 0^{\circ} + μg cos 3 0^{\circ} = \frac{5}{0.5} = 10$
$μ = \frac{1}{3} ≅ 0.6$