Question

Two boxes are connected to each other. The box on the table weighs 3kg, and on the other end is a suspended 1kg box which is one meter from the floor. The connecting string runs over a pulley. Without taking any friction into account, what is the kinetic energy of the 1kg box just before it hit the floor once the system is released from rest ????

Answer 1

force on 3kg box is 9.8N (because 1kg * 9.8m/s)

Answer 2

the acceleration the 3kg experiences is 9.8N/3kg = 3.26666 m/s^2

Answer 3

Vf ^2 = Vi^2 + 2ad

Answer 4

therefore the final velocity on the 3kg box Vf^2 = 0 + 2 (3.26666 m/s^2)(1m)

Answer 5

Vf = 2.5560 m/s

Answer 6

the small box experiences the same velocity since their connected by the string

Answer 7

KE = 1/2 mv^2
KE = 1/2 (1kg)(2.5560m/s)^2

Answer 8

Hi the answers given to me to choose from are: a) 2,45J b) 4,90J c) 9,80J d) 29,4J e) 39,2J

Answer 9

it should be a)
total PE=KE of 3kg mass + KE of 1kg mass
and since they have the same speed
1 x 1 x9.8 = 0.5 x 3 x v^2 + 0.5 x 1 x v^2
since they have the same speed.
solving for this you get v^2 = 4.9
KE of the 1kg box is 0.5mv^2=0.5 x 1 x4.9 = 2.95

Answer 10

According to the principle of conservation of energy, the loss of the gravitational is equal to the kinetic energy gained, thus
1/2 mv^ 2 = mgh, where 1/2 mv^2 is the formula to calculate kinetic energy while mgh is the formula to calculate potential energy.
Kinetic energy before hitting the ground = mgh
= (1)(10)(1) = 10 N

Answer 11

i am agree with @Matthewszh...used the principle of conservation of energy..
u may solve it like this...
and let the acceleration of the 1kg mass is downward...so the 3kg mass has the same acceleration...
for 1 kg mass-
1g-T=1*a
and for 3 kg mass-
T=3*a
from this two equation...a=g/4 [g=gravitational acceleration]
now let the velocity of the 1 kg mass just before hitting the ground be v
so v^2=0+2g/4*1=g/2
so the KE=1/2*1*v^2=g/4=2.45 J