Advertisements
Advertisements
Question
Figure shows (x, t), (y, t ) diagram of a particle moving in 2-dimensions.
|
|
 (b) |
If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.
Advertisements
Solution
Clearly, from the diagram (a), the variation can be related as x = t
⇒ `(dx)/(dt)` = 1 m/s
ax = 0
From diagram (b) y = t2
⇒ `(dy)/(dt)` = 2t or ay = `(d^2y)/(dt^2)` = 2 m/s2
Hence, Iy = may = 500 × 10–3 × 2 = 1 N .....(∵ m = 500 g)
Fx = max = 0
Hence, net force, F = `sqrt(F_x^2 + F_y^2)` = Fy = 1 N .....(along y-axis)
APPEARS IN
RELATED QUESTIONS
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms–1. How long does the body take to stop?
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.)
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
The figure shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region
(a) AB
(b) BC
(c) CD
(d) DE
A person says that he measured the acceleration of a particle to be non-zero even though no force was acting on the particle.
car moving at 40 km/hr is to be stopped by applying brakes in the next 4 m. If the car weighs 2000 kg, what average force must be applied to stop it?
In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 106 m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10−31 kg.
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.
In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m, so that it remains at rest?
Consider the situation shown in the following figure. Both the pulleys and the string are light and all the surfaces are frictionless.
- Find the acceleration of the mass M.
- Find the tension in the string.
- Calculate the force exerted by the clamp on the pulley A in the figure.
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s2.
An aeroplane is moving uniformly at a constant height under the action of two forces (i) Upward force (lift) and (ii) Downward force (weight). What is the net force on the aeroplane?
A body of mass m moving with a velocity v is acted upon by a force. Write an expression for change in momentum in each of the following cases: (i) When v << c, (ii) When v → c and (iii) When v << c but m does not remain constant. Here, c is the speed of light.
The unit of linear momentum is :
A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : The velocity acquired by the body
Why is it advantageous to turn before taking a long jump?
Use Newton's second law to explain the following:
We always prefer to land on sand instead of hard floor while taking a high jump.
A stone is dropped from a tower 98 m high. With what speed should a second stone be thrown 1 s later so that both hit the ground at the same time?
Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?
