Finding Friction's Magnitude
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On a rough horizontal surface, a body of mass 2 kg is given a velocity of 10 m/s. If the coefficient of friction is 0.2 and g=10 m/s2, the body will stop after covering a distance of
250 m
50 m
10 m
25 m
A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to the upper block. Take g = 10 m/s2.
a = 4 m/s, a = 4 m/s
a = 1 m/s, a = 4 m/s
None of these
a =2 m/s, a =2 m/s
Find the minimum force 'F' and the angle 'α' at which it should be applied to a block of mass 'm' kept on a horizontal surface such that the block starts moving. Given that the coefficient of friction between the block and surface is μ.
Fmin=μmg√1+μ2, α=tan−11μ
Fmin=μmg√1+μ2, α=tan−11μ
None of these
Fmin=μmg√1+μ2, α=tan−1μ
हैरी यह ज्ञात करने के लिए एक प्रयोग करता है, कि भिन्न-भिन्न पृष्ठ (A, B, C तथा D) 50 km/h से गतिशील एक कार को ब्रेक लगाकर रोकने के लिए आवश्यक दूरी को किस प्रकार प्रभावित करते हैं। परिणाम नीचे दी गई सारणी में दर्शाए गए हैं तथा किस प्रकार की सड़क कार के रूकने के लिए सर्वाधिक घर्षण प्रदान करेगी?
Types of road surface (सड़क के पृष्ठ के प्रकार) |
Stopping distance (m) (रूकने की दूरी (m)) |
A | 18 |
B | 25 |
C | 19 |
D | 17 |
- A
- B
- C
- D
A block of mass 'm' is supported on a rough wall by applying a force P as shown in figure. Coefficient of static friction between block and wall is μs
For what range of values of P, the block remains in static equilibrium?
None of these
- 2 m/s
- 4 m/s
- 6 m/s
- 8 m/s
A block released from rest from the top of a smooth inclined plane of inclination 45∘ 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∘ takes 2t seconds to reach the bottom. The coefficient of friction is
√0.5
√0.75
0.5
0.75
Figure shows a man standing stationary with respect to a horizontal conveyor belt. If the coefficient of static friction between the man's shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg).
- 33 N
- 13 N
- 10 N
- 20 N
For just moving the block along horizontal direction, identify the correct statement.
- If μ=0 , F is lesser in case (a)
- If μ=0 , F is lesser in case (b)
- If μ≠0, F is lesser in case (a)
- If μ≠0, F is lesser in case (b)
- 2mv2tanθℓ√4+tan2θ
- 2mv2tan2θℓ√2+tan2θ
- mv2tan2θℓ√4+tan2θ
- mv2tan2θℓ√2+tan2θ
- −^i N
- −18 ^i N
- −2.4 ^i N
- −3 ^i N
( Take tan 150=0.27, g=10ms−2)
- 10√3ms−1
- 9√10ms−1
- √10ms−1
- 2√10ms−1
- 26 N
- 19.5 N
- 10 N
- 30 N