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A body of mass 6kg rests on an inclined plane. The normal reaction R and the limiting frictional force is F as shown in the diagram (Fig. 2). If F is 30N and g=10ms-2 , then the angle of inclination Ө is
A. 15°
B. 60°
C. 45°
D. 30°
QUICK ANSWER…
D
DETAILS…
For an object on an inclined plane, the resolution of the weight W along the plane, opposite the frictional force (F) is = Wsinө
While the resolution of the weight perpendicular to the plane and opposite the normal reaction (R) is = Wcosө
But we are given the value of F and not R, this is why we will use:
F = Wsinө
But weight W = mass (m) x acceleration due to gravity (g),
Therefore, F = mg sinө
Sin ө = F/mg
= 30/(6 × 10)
= 30/60
= 1/2
= 0.5
Therefore, ө = sin¯¹0.5 = 30°
Now for the right answer to the above question:
- Option A is incorrect.
- Option B is incorrect.
- C is incorrect.
- D is the correct answer.
KEY-POINTS…
You may please note these/this:
- Resolution of the weight of an object along the slope of the inclined plane is given by F = mg Sinө
- Resolution of the same weight at 90 degrees to the plane is given by R = mg Cosө.
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/ culled from 2015 JAMB-UTME physics past question 4 /
