NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI05:LAWS OF MOTION
363395
A block of mass 2 \(kg\) is lying on a floor. The coefficient of static friction is 0.54. What will be the value of frictional force if the force is 2.8 \(N\) and \(g = 10\,m{s^{ - 2}}\)
363396
A block of mass 4 \(kg\) is placed on a rough horizontal plane. A time dependent force \(F = k{t^2}\) acts on the block, where \(k = 2N/{s^2}\). Coefficient of friction \(\mu = 0.8\). Force of friction between block and the plane at \(t = 2\,s\) is
1 \(8\,N\)
2 \(4\,N\)
3 \(2\,N\)
4 \(32\,N\)
Explanation:
\({f_{\max }} = \mu mg = 0.8 \times 4 \times 10 = 32N\) At \(t = 2s,F = k{t^2}(2){\left( 2 \right)^2} = 8N\) Since the applied force \(f < {f_{\max }}\), force of friction will be 8\(N\)
PHXI05:LAWS OF MOTION
363397
It is easier to pull a lawn roller than to push it because pulling
1 Involves sliding friction
2 Involves dry friction
3 Increases the effective weight
4 Decreases normal reaction
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363398
A 30 \(kg\) block rests on a rough horizontal surface. A force of 200 \(N\) is applied on the block. The block acquires a speed of 4 \(m\)/\(s\), starting from rest in 2 \(s\). What is the value of coefficient of friction?
363395
A block of mass 2 \(kg\) is lying on a floor. The coefficient of static friction is 0.54. What will be the value of frictional force if the force is 2.8 \(N\) and \(g = 10\,m{s^{ - 2}}\)
363396
A block of mass 4 \(kg\) is placed on a rough horizontal plane. A time dependent force \(F = k{t^2}\) acts on the block, where \(k = 2N/{s^2}\). Coefficient of friction \(\mu = 0.8\). Force of friction between block and the plane at \(t = 2\,s\) is
1 \(8\,N\)
2 \(4\,N\)
3 \(2\,N\)
4 \(32\,N\)
Explanation:
\({f_{\max }} = \mu mg = 0.8 \times 4 \times 10 = 32N\) At \(t = 2s,F = k{t^2}(2){\left( 2 \right)^2} = 8N\) Since the applied force \(f < {f_{\max }}\), force of friction will be 8\(N\)
PHXI05:LAWS OF MOTION
363397
It is easier to pull a lawn roller than to push it because pulling
1 Involves sliding friction
2 Involves dry friction
3 Increases the effective weight
4 Decreases normal reaction
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363398
A 30 \(kg\) block rests on a rough horizontal surface. A force of 200 \(N\) is applied on the block. The block acquires a speed of 4 \(m\)/\(s\), starting from rest in 2 \(s\). What is the value of coefficient of friction?
363395
A block of mass 2 \(kg\) is lying on a floor. The coefficient of static friction is 0.54. What will be the value of frictional force if the force is 2.8 \(N\) and \(g = 10\,m{s^{ - 2}}\)
363396
A block of mass 4 \(kg\) is placed on a rough horizontal plane. A time dependent force \(F = k{t^2}\) acts on the block, where \(k = 2N/{s^2}\). Coefficient of friction \(\mu = 0.8\). Force of friction between block and the plane at \(t = 2\,s\) is
1 \(8\,N\)
2 \(4\,N\)
3 \(2\,N\)
4 \(32\,N\)
Explanation:
\({f_{\max }} = \mu mg = 0.8 \times 4 \times 10 = 32N\) At \(t = 2s,F = k{t^2}(2){\left( 2 \right)^2} = 8N\) Since the applied force \(f < {f_{\max }}\), force of friction will be 8\(N\)
PHXI05:LAWS OF MOTION
363397
It is easier to pull a lawn roller than to push it because pulling
1 Involves sliding friction
2 Involves dry friction
3 Increases the effective weight
4 Decreases normal reaction
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363398
A 30 \(kg\) block rests on a rough horizontal surface. A force of 200 \(N\) is applied on the block. The block acquires a speed of 4 \(m\)/\(s\), starting from rest in 2 \(s\). What is the value of coefficient of friction?
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI05:LAWS OF MOTION
363395
A block of mass 2 \(kg\) is lying on a floor. The coefficient of static friction is 0.54. What will be the value of frictional force if the force is 2.8 \(N\) and \(g = 10\,m{s^{ - 2}}\)
363396
A block of mass 4 \(kg\) is placed on a rough horizontal plane. A time dependent force \(F = k{t^2}\) acts on the block, where \(k = 2N/{s^2}\). Coefficient of friction \(\mu = 0.8\). Force of friction between block and the plane at \(t = 2\,s\) is
1 \(8\,N\)
2 \(4\,N\)
3 \(2\,N\)
4 \(32\,N\)
Explanation:
\({f_{\max }} = \mu mg = 0.8 \times 4 \times 10 = 32N\) At \(t = 2s,F = k{t^2}(2){\left( 2 \right)^2} = 8N\) Since the applied force \(f < {f_{\max }}\), force of friction will be 8\(N\)
PHXI05:LAWS OF MOTION
363397
It is easier to pull a lawn roller than to push it because pulling
1 Involves sliding friction
2 Involves dry friction
3 Increases the effective weight
4 Decreases normal reaction
Explanation:
Conceptual Question
PHXI05:LAWS OF MOTION
363398
A 30 \(kg\) block rests on a rough horizontal surface. A force of 200 \(N\) is applied on the block. The block acquires a speed of 4 \(m\)/\(s\), starting from rest in 2 \(s\). What is the value of coefficient of friction?