Newton’s Second Law corrected the old belief that force is needed to maintain motion. This idea came from the philosopher ______.
Advertisements
Advertisements
प्रश्न
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.
बेरीज
Advertisements
उत्तर
a = 3.26 m/s2, T = 3.9 N
After 2 s, velocity of mass m1,
v = u + at = 0 + 3.26 × 2
= 6.52 m/s upward
At this time, m2 is moving 6.52 m/s downward.
At time 2 s, m2 stops for a moment. But m1 is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.
Here, v = 0, u = 6.52 m/s
a = −g = − 9.8 m/s2
v = u + at = 6.52 + (−9.8)t
\[\Rightarrow t = \frac{6 . 52}{9 . 8} \approx \frac{2}{3} \sec\]
After this time, the mass m1 also starts moving downward.
So, the string becomes tight again after \[\frac{2}{3} s\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
The below figure shows the position-time graph of a particle of mass 4 kg.
- What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
- What is the impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)
An object is placed far away from all the objects that can exert force on it. A frame of reference is constructed by taking the origin and axes fixed in this object. Will the frame be necessarily inertial?
If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be
(a) going up with increasing speed
(b) going down with increasing speed
(c) going up with uniform speed
(d) going down with uniform speed
A person says that he measured the acceleration of a particle to be non-zero even though no force was acting on the particle.
A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg, respectively. Assuming that the magnitudes of acceleration and deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s2.
Find the mass M of the hanging block in the following figure that will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
Use Newton's second law of motion to explain the following instance :
An athlete prefers to land on sand instead of hard floor while taking a high jump .
Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m s-2.
A car is moving with a uniform velocity 30 ms-1. It is stopped in 2 s by applying a force of 1500 N through its brakes. Calculate the following values : The change in momentum of car.
State the magnitude and direction of the force of gravity acting on the body of mass 5 kg. Take g = 9.8 m s-2.
Name the physical entity used for quantifying the motion of a body.
Why is it advantageous to turn before taking a long jump?
State Newton's second law of motion.
A body of mass 400 g is resting on a frictionless table. Find the acceleration of the body when acted upon by a force of 0.02 N.
A stone is dropped from a cliff 98 m high.
How long will it take to fall to the foot of the cliff?
What do you mean by the conservation of momentum? Briefly, explain the collision between two bodies and the conservation of momentum.
A metre scale is moving with uniform velocity. This implies ______.
The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?
- The force at t = (1/8) s on the particle is – 16π2 Am.
- The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.
- The particle is not acted upon by any force.
- The particle is not acted upon by a constant force.
- There is no impulse acting on the particle.
Why is catching a slow-moving ball easier than catching a fast-moving ball?
