# Mechanics- Dynamics of particle moving in a straight line questions

• cakeislife
In summary: Also note that in the absence of any other information, the passengers are not accelerating relative to the lift. They are accelerating relative to the ground. (with the same acceleration).
cakeislife

## Homework Statement

1. A small pebble of mass 50g is dropped into a pond and falls vertically through it with an acceleration of 2.8ms^-2. Assuming that the water produces a constant resistance, find its magnitude.

2. A lift of mass 500kg is lowered or raised by means of a metal cable attached to its top. The lift contains passengers whose total mass is 300kg. The lift starts from rest and accelerates at a constant rate, reaching a speed of 3ms^-1 after moving a distance of 5m. Find:

a) The acceleration of the lift
b)the tension in the cable if the lift is moving vertically downwards
c)the tension in the cable if the lift is moving vertically downwards

3. A trailer of mass 200kg is attached to a car by a light tow bar. The trailer is moving along a straight horizontal road and decelerates at a constant rate from a speed of 15ms^-1 to a speed of 5ms^-1 in a distance of 25m. Assuming there is no resistance to the motion, find

a) the deceleration of the trailer
b) the thrust in the tow bar

4. A woman of mass 60kg is in a lift which is accelerating upwards at a rate of 2ms^-2
a) Find the magnitude of the normal reaction of the floor of the lift on the woman
The lift then moves at a constant speed and then finally decelerate to rest at 1.5ms^-2
b) Find the magnitude of the normal reaction of the floor of the lift on the woman during the period of deceleration

5. Albert and Bella are both standing in a lift. The mass of the lift is 250kg. As the lift moves upward with constant acceleration, the floor of the lift exerts forces of magnitude 678N and 452N respectively on Albert and bella. The tension in the cable which is pulling the lift upwards is 3955N
a) Find the acceleration of the lift
b) Find the mass of Albert
c) Find the mass of Bella

F=ma
SUVAT equations

## The Attempt at a Solution

1. I got the right answer for this question, however I don't really understand what thrust is?! :S I was meant to draw a diagram and I wasn't sure in which direction it would be acting, I presumed it's meant to be upwards on the stone?

2a)This was fine I used v^2=u^2 + 2as and got 0.9ms^-2 as the acceleration
b and c) Ok I don't understand why the reaction force of the passengers is not taken into consideration when finding the tension. Surely this would contribute to the force acting upwards?! Since the lift is accelerating, surely R is not equal to the weight of the passengers?! :S

3a) Again a was fine I used v^2=u^2 + 2as and I got -4ms^-1
b) I got the right answer, but again what is thrust?! Which way is it acting, on the trailer? The car? :S It is equal to the thrust acting on the car?

4a) No problems, used F=ma and got R=708N
b)Right, so I got the wrong answer here, this is my working out, not sure why my answer is wrong, in order for it to be right, acceleration has to be positive?! :S
W-R=ma
60g-R=60x(-1.5)
(60x9.8)-R= -90
R=588+90
=678N--> which is wrong

5a) Right so I worked this out, got the wrong acceleration, then saw the right value of the acceleration and managed to work it out, but I don't understand why it is that? They made the reaction forces equal to the weight of Albert and Bella respectively so that...
T-W=ma
3955-[(250x9.8) + 678 + 472]=250a
...eventually getting a=1.5ms^-2
But surely the reaction forces can't be equal to the weight, because then that would mean in b) and c) you would get the wrong answer? :s

Thank you!

cakeislife said:
1. I got the right answer for this question, however I don't really understand what thrust is?!
Thrust is just force, usually in the context of a pushing force.
I was meant to draw a diagram and I wasn't sure in which direction it would be acting, I presumed it's meant to be upwards on the stone?
yes
b and c) Ok I don't understand why the reaction force of the passengers is not taken into consideration when finding the tension. Surely this would contribute to the force acting upwards?! Since the lift is accelerating, surely R is not equal to the weight of the passengers?! :S
b) and c) in the OP are identical. I guess one should be upwards. Neither specifies whether this is during acceleration or at constant speed.
3a) Again a was fine I used v^2=u^2 + 2as and I got -4ms^-1
b) I got the right answer, but again what is thrust?! Which way is it acting, on the trailer? The car? :S It is equal to the thrust acting on the car?
which way is the car pushing the trailer while decelerating? Which way is the trailer pushing the car?
4a) No problems, used F=ma and got R=708N
b)Right, so I got the wrong answer here, this is my working out, not sure why my answer is wrong, in order for it to be right, acceleration has to be positive?! :S
W-R=ma
60g-R=60x(-1.5)
(60x9.8)-R= -90
need to be consistent with signs. ∑ means add up the forces: mg+R. If taking up as positive, a is negative and so is g.
5a) Right so I worked this out, got the wrong acceleration, then saw the right value of the acceleration and managed to work it out, but I don't understand why it is that? They made the reaction forces equal to the weight of Albert and Bella respectively so that...
T-W=ma
3955-[(250x9.8) + 678 + 472]=250a
...eventually getting a=1.5ms^-2
But surely the reaction forces can't be equal to the weight, because then that would mean in b) and c) you would get the wrong answer? :s

Thank you!
The 678 and 472 are really of the form m(-g+a), g being negative. If you rearrange the equation to get all the 'a' terms on the right, you will see it conforms to T-W = ma.

## 1. What is the equation for calculating the velocity of a particle moving in a straight line?

The equation for velocity is v = d/t, where v is the velocity, d is the distance traveled, and t is the time taken.

## 2. How do you calculate the acceleration of a particle in a straight line?

The formula for acceleration is a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken.

## 3. Can the velocity of a particle change without changing its acceleration?

Yes, the velocity of a particle can change without changing its acceleration if the direction of the particle's velocity changes while the magnitude of its acceleration remains constant.

## 4. How does mass affect the acceleration of a particle in a straight line?

According to Newton's Second Law of Motion, the acceleration of a particle is directly proportional to the net force acting on the particle and inversely proportional to its mass. This means that as mass increases, the acceleration decreases and vice versa.

## 5. What is the difference between uniform and non-uniform motion?

Uniform motion is when a particle moves in a straight line with a constant velocity, while non-uniform motion is when the velocity of the particle changes over time. In other words, uniform motion has no acceleration, while non-uniform motion has a non-zero acceleration.

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