Question

Mechanics Question 5

Answer 1

@Abhisar

Answer 2

Hi !
Hv u tried it ?

Answer 3

#1
Work done = Frictional force X Distance

Answer 4

For #1
Work done by B against the frictional force = Loss in potential energy of the system = Gain in K.E of the system.

Answer 5

R=0.3g=3N
Ff=μR=0.7*3=2.1N
Work done=Frictional Force*Distance moved=2.1*0.9=1.9 J

Answer 6

Potential energy of mass B is 0

Answer 7

ok, so you got part a
for part b, the potential energy of mass B doesn't change, so we can ignore it
the change in potential energy of mass A will depend on it's change in height

Answer 8

should we use the initial height of A as 0.9

Answer 9

if so why?

Answer 10

well it doesn't really matter where we take our origin for height, we know that the CHANGE in height is 0.9 meters, and that's all we need

Answer 11

how is the change in height 0.9

Answer 12

it gives you that in the question, that's how far the thing moves

Answer 13

P.E=mg(hf-hi)=10*0.3*(-0.9)=-2.7 J this represent the loss of P.E of the system.
Gain in K.E= Loss of P.E+Work done against Resistance
=2.7-1.9

Answer 14

well you've got the numbers right i think, but you meant to say
gain in ke = loss of pe - work done against resistance, maybe ?

Answer 15

yes i am sorry for that

Answer 16

shall i leave you to think about the last part ?

Answer 17

i think we should find the acceleration for the first stage when particle A moved 0.54 m to find the initial speed.

Answer 18

there's an easier way

Answer 19

you just calculated the kinetic energy of the two mass system, and they are of equal mass moving at the same speed, so each has half of the kinetic energy

Answer 20

hence you can find the speed of mass A when the string breaks

Answer 21

1/2mv^2=0.45
1/2(0.3)^2=0.45
v=(3)^1/2

Answer 22

I made it square root of 2.7

Answer 23

but your reasoning is correct

Answer 24

now use s=0.54 m a=10 m/s^2 as its not anymore attached to the string.

Answer 25

correct

Answer 26

use v^2=u^2+2as

Answer 27

yes, that's right

Answer 28

oh thanks i got it now

Answer 29

you're welcome