The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination - - 45-tan- from bottom to top and from top to bottom is 5- The coefficient of friction between the block and surface is-

The correct option is B 2/3Apply work energy theorem #160; W_ B T + W _ mg + W _ f =0 W_ B T mgsinθ× 5 μ mgcosθ× 5=0 W_ B T =( mgsinθ +μ mgc ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Centripetal Acceleration

The ratio of ...

Question

The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination $θ$ = $45^{\circ} (t a n θ > μ)$ from bottom to top and from top to bottom is 5. The coefficient of friction between the block and surface is:

1/2

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

2/3

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

2/5

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

4/5

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

Open in App

Solution

The correct option is B 2/3
Apply work energy theorem
$W_{B - T} + W_{m g} + W_{f} = 0 W_{B - T} - m g s i n θ \times 5 - μ m g c o s θ \times 5 = 0 W_{B - T} = (m g s i n θ + μ m g c o s θ) 5 S i m i l a r l y W_{B - T} + m g s i n θ \times 5 - μ m g c o s θ \times 5 = 0 W_{B - T} = (m g s i n θ - μ m g c o s θ) 5 \frac{(m g s i n θ + μ m g c o s θ)}{- (m g s i n θ + μ m g c o s θ)} = 5 4 m g s i n θ = 6 μ m g c o s θ μ = \frac{2}{3}$

Suggest Corrections

Similar questions

Q. A block of mass m is released from top of incline plane of inclination. The coefficient of friction between the block and incline surface is

μ (μ < t a n θ)

. Find work done by normal reaction when block moves from top to the bottom.

Q. A block released from rest from the top of a smooth inclined plane of inclination

45^{\circ}

, takes time

t

to reach the bottom. The same block released from rest, from top of a rough inclined plane of same inclination takes time

2 t

to reach the bottom. The coefficient of friction is

A block released from rest from the top of a smooth inclined plane of inclination $45^{\circ}$ takes $t$ seconds to reach the bottom. The same block released from rest from the top of a rough inclined plane of the same inclination of $45^{\circ}$ takes $2 t$ seconds to reach the bottom. The coefficient of friction is

Q. A block of mass m is released from the top of a rough inclined plane as shown in the figure. Coefficient of friction between the block and the plane is

μ

. Speed of the block at the bottom of the plane is v.

Join BYJU'S Learning Program

The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination θ = 45∘(tanθ>μ) from bottom to top and from top to bottom is 5. The coefficient of friction between the block and surface is:

The correct option is B 2/3Apply work energy theorem WB−T+Wmg+Wf=0WB−T−mgsinθ×5−μmgcosθ×5=0WB−T=(mgsinθ+μmgcosθ)5SimilarlyWB−T+mgsinθ×5−μmgcosθ×5=0WB−T=(mgsinθ−μmgcosθ)5(mgsinθ+μmgcosθ)−(mgsinθ+μmgcosθ)=54mgsinθ=6μmgcosθμ=23

The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination $θ$ = $45^{\circ} (t a n θ > μ)$ from bottom to top and from top to bottom is 5. The coefficient of friction between the block and surface is: