JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 14)
An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile.
_1st_September_Evening_Shift_en_14_1.png)
The correct applied force vs distance graph will be :
The correct applied force vs distance graph will be :
_1st_September_Evening_Shift_en_14_2.png)
_1st_September_Evening_Shift_en_14_3.png)
_1st_September_Evening_Shift_en_14_4.png)
_1st_September_Evening_Shift_en_14_5.png)
Explanation
During upward motion
F = 2N = (+ve) constant
_1st_September_Evening_Shift_en_14_6.png)
During downward motion
$$\Rightarrow$$ F = 2N = ($$-$$ve) constant
_1st_September_Evening_Shift_en_14_7.png)
$$\Rightarrow$$ Best possible answer is option (c)
F = 2N = (+ve) constant
During downward motion
$$\Rightarrow$$ F = 2N = ($$-$$ve) constant
$$\Rightarrow$$ Best possible answer is option (c)
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