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
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 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
Acceleration of the block is :
Horizontal component of force is F cosθ = 20√2 cos 45° = 20√2 × 1/√2 = 20 N.
So, Horizontal acceleration = 20N / 10kg = 2m/s²
A hammer of man 1 kg moving with a speed of 6 m/sec strikes a wall and comes to rest in 0.1 sec find average retarding force that stops the hammer :
According to Newtons second law of motion, Force is time rate of change in momentum.
So, F = change in momentum / time
F= m (vf-vi)/t= 1 kg (0-6)/0.1= -60 N
A 5 kg block is resting on a frictionless plane, it is struck by a Jet, releasing water at the rate of 4 kg/sec emerging with a speed of 5 m/sec. Calculate the initial acceleration of block :
Force acting on the object is F = v.(dm/dt)= 5×4 = 20 N
so Acceleration is a= F/m = 20/5 = 4 m/s²
A body of mass 1 kg is acted upon by two perpendicular forces of 8 N and 6 N, find the magnitude of acceleration :
Resultant force on the object will be F= (8² + 6²)½= 10 N
So, acceleration is a = F/m = 10/1 = 10 m/s²
A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn/Sn+1 is :
Sn = u + a/2(2n-1) = a/2(2n-1)
Sn+1 = u + a/2(2(n+1)-1) = a/2(2n+1)
Sn / Sn+1 = (2n-1)/(2n+1)