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A small block of mass m is kept on a rough inclined surface of inclination q fixed in a elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be :
- a)zero
- b)mgvt cos2θ
- c)mgvt sin2θ
- d)mgvt sin 2θ
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A small block of mass m is kept on a rough inclined surface of inclina...
The work done by the force of friction can be calculated using the formula:
Work = Force x Distance
In this case, the force of friction is opposing the motion of the block, so it is in the opposite direction of the velocity of the elevator. The force of friction is given by:
Force of friction = coefficient of friction x normal force
The normal force can be calculated by resolving the weight of the block into components perpendicular and parallel to the inclined surface. The perpendicular component is given by:
Normal force = mg cos(q)
The parallel component is given by:
Force parallel = mg sin(q)
Since the block does not slide on the wedge, the force of friction must be equal to the parallel component of the weight:
Force of friction = mg sin(q)
The distance traveled by the block in time t can be calculated using the velocity and time:
Distance = velocity x time
Therefore, the work done by the force of friction is:
Work = Force of friction x Distance
= (mg sin(q)) x (velocity x time)
= mgvt sin(q) x t
So, the work done by the force of friction on the block in time t is mgvt sin(q) x t. Therefore, the correct option is not provided in the given answer choices.
Work = Force x Distance
In this case, the force of friction is opposing the motion of the block, so it is in the opposite direction of the velocity of the elevator. The force of friction is given by:
Force of friction = coefficient of friction x normal force
The normal force can be calculated by resolving the weight of the block into components perpendicular and parallel to the inclined surface. The perpendicular component is given by:
Normal force = mg cos(q)
The parallel component is given by:
Force parallel = mg sin(q)
Since the block does not slide on the wedge, the force of friction must be equal to the parallel component of the weight:
Force of friction = mg sin(q)
The distance traveled by the block in time t can be calculated using the velocity and time:
Distance = velocity x time
Therefore, the work done by the force of friction is:
Work = Force of friction x Distance
= (mg sin(q)) x (velocity x time)
= mgvt sin(q) x t
So, the work done by the force of friction on the block in time t is mgvt sin(q) x t. Therefore, the correct option is not provided in the given answer choices.
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A small block of mass m is kept on a rough inclined surface of inclination q fixed in a elevator.The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be :a)zerob)mgvt cos2θc)mgvt sin2θd)mgvt sin 2θCorrect answer is option 'C'. Can you explain this answer?
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