## A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30° with horizontal) with constant velocity ….

Q: A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30° with horizontal) with constant velocity of 5 m/s as shown in the figure. Find the friction force acting on a body by the inclined plane. ( take g = 10 m/s2 )

(A) 25 N

(B) 20 N

(C) 30 N

(D) none of these

Click to See Solution :
Ans: (A)
Sol: Since the block slides down the incline with uniform
velocity, net force on it must be zero. Hence mg sinθ
must balance the frictional force ‘ f ‘ on the block.
Therefore f = mg sinθ = 5 × 10 × (1/2) = 25 N

## A body of mass 10 kg lies on a rough inclined plane of inclination θ = sin-(3/5) with the horizontal….

Q: A body of mass 10 kg lies on a rough inclined plane of inclination θ = sin(3/5) with the horizontal. When a force of 30 N is applied on the block parallel to & upward the plane, the total reaction by the plane on the block is nearly along

(A) OA

(B) OB

(C) OC

(D) OD

Click to See Solution :
Ans: (A)
Sol: Frictional force along the in upward direction = 10 g sinθ – 30 = 30 N
N = 10 g cosθ = 80 NDirection of R is along OA

## A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction between…

Q: A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction between A & B is 0.5. (t is time in seconds. Initial velocities are zero, A is always on B)

(A) block A starts sliding on B at t = 5 seconds

(B) the heat produced due to friction in first 5 seconds is 312.5 J

(C) the heat produced due to friction in first 5 seconds is (625/8) J

(D) acceleration of A at 10 seconds is 5 m/s

Click to See Solution :
Ans: (A) , (D)
Sol:

fmax = μ × 3 g = 0.5 × 30 = 15 N

block A starts sliding when friction force becomes max. i.e. fmax = 15 at that instant (F.B. D.)

both will move with same acceleration
So , 15 = 3a

a = 5 m/s2
F – 15 = 7 a

10 t – 15 = 7 × 5

10 t = 50

t = 5 sec
Work done by friction in 5 seconds

$W = \int F.dS = \int 10 t .dS$

$W = \int_{0}^{5} 10 t .v dt$ ;Since $v = \int a dt = \int t dt$ & a = F/m = 10t/10 = t ;

$W = \int_{0}^{5} 10 t .\frac{t^2}{2} dt$

$W = \int_{0}^{5} 5 t^3 dt$

$W = 5[\frac{t^4}{4}]_{0}^{5}$

$W = \frac{625 \times 5}{4} J$

## In the shown arrangement if f1, f2 and T be the frictional forces on 2 kg block, 3 kg block and tension in the string …

Q: In the shown arrangement if f1, f2 and T be the frictional forces on 2 kg block, 3 kg block and tension in the string respectively, then their values are:

(A) 2 N , 6 N , 3.2 N

(B) 2 N , 6 N, 0 N

(C) 1 N , 6 N , 2 N

(D) data insufficient to calculate the required values

Click to See Solution :
Ans: (C)

Sol:

Net force without friction on system is 7 N in right
side so first maximum friction will come on 3 kg block.

So f2 = 1 N, f3 = 6 N, T = 2