When a constant force is applied to a body, it moves with uniform
From Newton's second law of motion, F= m.a
if force is constant acceleration remains constant.
A particle is moving with a constant speed along a straight line path. A force is not required to
As the particle moves with constant speed in straight line its velocity is constant i.e accleration is zero. From Newton's second law of motion
F=m.a
Force required to keep moving with uniform velocity is zero.
A fireman wants to slide down a rope. The rope can bear a tension of (3/4)th of the weight of the man. With what minimum acceleration should the fireman slide down:
According to the question, T=3mg/4
Using equations of motion, i.e. Fnet = m.a
⇒mg-T =ma
⇒mg-3mg/4= ma
⇒a =g/4
Consider the following two statements:
(A) Linear momentum of a system of particles is zero
(B) Kinetic energy of a system of particles is zero
Then:
When kinetic energy of the system is zero, speed of objects will be zero so momentum will always be zero.
Whereas momentum being a vector quantity may be zero when directions of velocities cancel each other. so speed and kinetic energy may of may not be zero when momentum is zero
The linear momentum of a particle is given by P = a + bt², where t is time and a & b are constants. The force acting on the body is directly proportional to :
Given momentum of the object is P = a + bt²;
Force is defined as rate of change of momentum F= dP/dt= 2bt. so Force is proportional to t.
A body of mass 1 kg moving on a smooth horizontal surface with velocity 5m/s is to be brought to rest in 5 sec. How much force need to be applied :
Force is defined as rate of change of momentum ie. F= dP/dt= change in momentum / time
Here Force = (mv-mu)/t= m(v-u)/t= 1 kg (0- 5)/ 5 = 1 N
The momentum of a particle is P = 2cos tˆi + 2sin t ˆj. What is the angle between the force F acting on the particle and the momentum P ?
Momentum of object is given as, P = 2cos tˆi + 2sin t ˆj ,
From Newtons Second law, F = dP/dt = -2 sin t î + 2 cos t ˆj.
Taking Dot product of P and F we get = -4 sint . cos t + 4 sint .cost =0 .
So, force and momentum are perpendicular to each other.
Two forces of 6N and 8N act on a body of mass 7 kg. The value of acceleration produced can not be more than :
Maximum resultant force will act on the object when 6N and 8 N forces will be acting in same direction. F max= 14 N.
Maximum Acceleration= Maximum Force/ mass= 14 N/ 7 kg = 2m/s²
A body moving with uniform velocity is stopped in 0.25 sec by applying a force of 200 N. What is the impulse :
Impulse is change in momentum so, J = F.dt = 200 × 0.25 =50 N-s
A rocket of mass 6000 kg is set for vertical firing. If the exhaust speed be 1 km/s how much gas must be ejected to give the rocket an upward acceleration of 20 m/s² (neglect gravity) :-
According to Newtons second law of motion
F= v. dm/dt
so, acceleration = v. (dm/dt)/m
⇒ 20 m/s² = 1000 ×(dm/dt)/6000
⇒dm/dt = 120 kg/sec