Calculating Friction Using Coeffcients
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(Take g=10 ms−2)
- 2.96 s
- 2 s
- 1.75 s
- 3.14 s
(Given μ=13 and g=10 m/s2)
- 50 N along +x direction
- 100 N along +x direction
- 50 N along −x direction
- 100 N along −x direction
- 80 N
- 120 N
- 150 N
- 100 N
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
- if μ1=0.5 and μ2=0.3, then 5 kg block exerts 3 N force on the 3 kg block
- if μ1=0.5 and μ2=0.3, then 5 kg block exerts 8 N force on the 3 kg block
- if μ1=0.3 and μ2=0.5, then 5 kg block exerts 1 N force on the 3 kg block
- if μ1=0.3 and μ2=0.5, then 5 kg block exerts no force on the 3 kg block
- cos θμ0
- sin θμ0
- tan θμ0
- 2 tan θμ0
Given μs=0.5 and μk=0.4. Taking g=10 m/s2, find the acceleration in each case.
- zero, zero
- zero, 5 m/s2
- zero, 6 m/s2
- 4 m/s2, 6 m/s2,
(Assume there is no slipping between the blocks).
- 53 N
- 103 N
- 2 N
- 0 N
- The coefficient of friction between the block and the table is 0.1
- The coefficient of friction between the block and the table is 0.2
- If the table has half of its present roughness, the time taken by the block to complete the journey is 4 s
- If the table has half of its present roughness, the time taken by the block to complete the journey is 8 s
- 0.2
- 0.3
- 0.4
- 0.8
- 4g3
- g3
- g2
- 3g4
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
- 25
- 35
- 34
- 58
- 4g3
- g3
- g2
- 3g4
. (take g=10 ms−2)
A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N of the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is
2.5 N
0.98 N
4.9 N
0.49 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
- 0.2
- 0.3
- 0.4
- 0.8
- 0.35
- 0.45
- 0.2
- 5
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
- tanθ≥μ
- cotθ≥μ
- tanθ2≥μ
- cotθ2≥μ
- 2 tan θ
- tan θ
- 2 sin θ
- 2 cos θ
(Given m1=4 kg, m2=2 kg, μ1=0.8, μ2=0.5, g=10 m/s2, sin37∘=3/5)
- 3.2 N
- 3.6 N
- 7.2 N
- Zero
- 2.5 s
- 5 s
- 7.5 s
- 10 s
- tanθ≥μ
- cotθ≥μ
- tanθ2≥μ
- cotθ2≥μ