Calculating Friction Using Coeffcients
Trending Questions
- 0.35
- 0.45
- 0.2
- 5
- 2 m
- 4 m
- 3 m
- 1 m
- False
- True
Two blocks of masses m1=4 kg and m2=6 kg are connected by a string of negligible mass passing over a frictionless pulley as shown in the figure. The coefficient of friction between block m1 and the horizontal surface is 0.4. When the system is released, the masses m1 and m2 start accelerating. What additional mass m should be placed over mass m1 so that the masses (m1+m) slide with a uniform speed?
9 kg
10 kg
11 kg
12 kg
- 20 N
- 30 N
- 10 N
- Zero
- 9.84m/s
- 10.84m/s
- 7.84m/s
- 5.84m/s
- 4 m/s2, T=0
- 2 m/s2, T=5 N
- 10 m/s2, T=10 N
- 15 m/s2, T=9 N
- 0.33
- 0.75
- 0.80
- 0.25
- −1 m/s2
- 1.2 m/s2
- 0.2 m/s2
- zero
- (M+m)(μ2−μ1)g
- (M−m)(μ2−μ1)g
- (M−m)(μ2+μ1)g
- (M+m)(μ2+μ1)g
A shell of mass m at rest explodes in two parts with mass ratio 1:2. The surface has friction coefficient μ. What is the ratio of stopping distance
4/1
3/2
3/1
2/1
- 2 tanθ
- tanθ
- 2 cotθ
- 2 cosθ
A boy of mass m is sliding down a vertical pole by pressing it with a horizontal force f. If μ is the coefficient of friction between his palms and the pole, the acceleration with which he slides down will be equal to
g
μfm
g+μfm
g−μfm
A trolley of mass M is attached to a block of mass m by a string passing over a frictionless pulley as shown in the figure. If the coefficient of friction between the trolley and the surface below is μ, what is the acceleration of the trolley and the block system when they are released?
(m−Mm+M)g
mMg
(μm−Mm+M)g
(m−μMm+M)g
- 0.20
- 0.35
- 0.25
- 0.15
- −1 m/s2
- 1.2 m/s2
- 0.2 m/s2
- zero
- 10 N
- 20 N
- 2 N
- 1 N
- μ>ag
- μ<ga
- μ<ag
- μ>ga
Two blocks A and B are connected to each other by a string and a spring of force constant k, as shown in the figure. The string passes over a frictionless pulley. Block B slides over the horizontal top surface of a stationary block C and block A slides along the vertical side of C both with the same uniform speed. If the coefficient of friction between the surfaces of the blocks is μ and the mass of block A is m, what is the mass of block B?
m√μ
mμ
√μm
μm
(Take g=10 ms−2)
- 2.96 s
- 2 s
- 1.75 s
- 3.14 s
- 25 N
- 75 N
- 50 N
- 100 N
The block m1 is loaded on the block m2. If there is no relative sliding between the blocks and the inclined plane is smooth, find the friction force between the blocks.
m1gsinθ
(m1+m2)g sinθ
0
μm1gcosθ
- v=[rg(tanθ−μ1+μtanθ)]
- v=[rg(tanθ+μ1−μtanθ)]1/2
- v=[rg(tanθ+μ1+μtanθ)]1/2
- v=[rg(tanθ−μ1+μtanθ)]1/2
- the force of gravity disappears
- he loses weight some how
- he is kept in this path due to the force exerted by surrounding air
- the frictional force of the wall balances his weight
(All pulleys are light and frictionless, string is massless and inextensible)
(Take g=10 m/s2)
- 0
- 5 m/s2
- 10 m/s2
- 4 m/s2