A body of mass 100 g is sliding from an inclined plane of inclination 30° . What is the frictional force experienced if μ =0.7 ?

Question

A body of mass 100 g is sliding from an inclined plane of inclination 30° . What is the frictional force experienced if μ =0.7 ? (A)  0.7× √2× (1/√3)N  (B)  0.7× √3× (1/2)N  (C)  0.7× √3N  (D)  0.7× 2× (1/√3)N

Tardigrade Admin · Accepted Answer

: F=μ R=μ mg cos θ F=0.7× (100/1000)× 10× cos 30° =0.7× (√3/2)N

Q. A body of mass 100 g is sliding from an inclined plane of inclination $30^{\circ}$ . What is the frictional force experienced if $μ = 0.7$ ?

$0.7 \times 2 \times \frac{1}{3} N$

$0.7 \times 3 \times \frac{1}{2} N$

$0.7 \times 3 N$

$0.7 \times 2 \times \frac{1}{3} N$

: $F = μ R = μ m g cos θ$ $F = 0.7 \times \frac{100}{1000} \times 10 \times cos 30^{\circ} = 0.7 \times \frac{3}{2} N$

Q. A body of mass 100 g is sliding from an inclined plane of inclination 30∘ . What is the frictional force experienced if μ=0.7 ?

0.7×2​×3​1​N

0.7×3​×21​N

0.7×3​N

0.7×2×3​1​N

: F=μR=μmgcosθ F=0.7×1000100​×10×cos30∘=0.7×23​​N

Q. A body of mass 100 g is sliding from an inclined plane of inclination $30^{\circ}$ . What is the frictional force experienced if $μ = 0.7$ ?

$0.7 \times 2 \times \frac{1}{3} N$

$0.7 \times 3 \times \frac{1}{2} N$

$0.7 \times 3 N$

$0.7 \times 2 \times \frac{1}{3} N$

: $F = μ R = μ m g cos θ$ $F = 0.7 \times \frac{100}{1000} \times 10 \times cos 30^{\circ} = 0.7 \times \frac{3}{2} N$