Three blocks A, B, and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a 14 N is applied to the 4 kg block, then the contact force between A and B is
2 N
6 N
8 N
8 N
B.
6 N
Given, mA = 4 kg
mB = 2 kg
=> mC =1 kg
So total mass (M) = 4+2+1 = 7 kg
Now, F = Ma
14 = 7a
a=2 m/s2
F-F' = 4a
F' = 14-4x2
F' = 6N
A car is negotiating a curved road of radius R. The road is banked at angle θ. The coefficient of friction between the tyres of the car and the road is μs . The maximum safe velocity on this road is,
A.
A car is negotiating a curved road of radius R. The road is banked at angle and the coefficient of friction between the tyres of car and the road is .
The given situation is illustrated as:
In the case of vertical equilibrium,
N cos = mg + f1 sin mg = N cos ... (i)
In the case of horizontal equilibrium,
... (ii)
Dividing Eqns. (i) and (ii), we get
A particle of mass m is driven by a machine that delivers a constant power K watts. If the particle starts from rest, the force on the particle at time t is
A.
As the machine delivers a constant power
So F, v =constant = k (watts)
The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is
120°
60°
90°
150°
A.
120°
F1 = (F) one force
F2 = (2F) second force
Let θ is angle between F1 and F2
Here angle made by resultant FR is α with F1
A car moving with a velocity of 20 ms-1 stopped at a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is
320 m
1280 m
160 m
640 m
C.
160 m
Velocity of car (u) = 20 ms-1
Distance (s) = 40 m
From Newton's third equation,
In the second condition, the velocity becomes twice i.e, u' = 2u
Again from Newton's third equation, we get