a body of mass 5 kg is pulled 4 m up a rough plane inclined at 30° to the horizontal by a force of 50N parallel to the plane find the velocity of the block when it reaches to the top of the plane if the frictional force is of magnitude 12 N
Answers
Here angle of repose α=tan
−1
(μ
s
)
α=tan
−1
(
2
1
)=30
0
The angle of inclination is greater than the angle of respose. The friction force on the block will act in upward direction.
For the acceleration of block,
Mgsinθ−μN=Ma
⇒a=gsinθ−μgcosθ
=10(sin37
0
−0.5cos37
0
)
b. If external force F= 75N is applied on the block.
Let us find net driving force acting on block. Parallel to inclined two external forces are acting one in upward direction F and other is the component of weight in the direction downward the plane, Mgsinθ.
Net driving force f
driving
=F−Mgsinθ
⇒F
driving
=75−10×10×sin37
0
=75−60=15N
Maximum resisting force that oppose relative motion is maximum friction force (or f
lim
)
f
lim
=μ
s
Mgcosθ
=0.5×10×10×cos37
0
=40N
Here f
lim
<f
resisting
. Hence, the block will not move and friction will be static will act in the direction opposite to driving force, i.e., in downward direction.
c. To move block, the least value of driving force should be 40 N. But in above case, driving force is 15N (up). Hence, if we add △F=25N in upward direction, the block will overcome maximum resistence force (or friction) and starts moving up.
∴60+40=75+△F⇒△F=25N
d. As resisting force which is maximum friction force is 40N and the component of weight parallel to incline is 60 N and acting downward. If we remove F, then the driving forces will be the only component of the weight in the direction downward the incline plane. In this case, friction will act in upward direction. Hence, the required value of F to make block in equilibrium,
F + 40 = 60
or F = 20N