Question

Two boxes are connected to each other as shown. The system is released from rest and the 1.00-kg box falls through a distance of 1.00 m. The surface of the table is frictionless. What is the kinetic energy of box b just before it reaches the floor?

Answer 1

Answer:

The kinetic energy of box b is 2.44 J.

Explanation:

Given that,

Mass of box a =3.00 kg

Mass of box b=1.00 kg

Distance = 1.00 m

We need to calculate the acceleration of the system

Using formula of acceleration

$a=\dfrac{m_{2}\times g}{m_{1}+m_{2}}$

put the value into the formula

$a=\dfrac{1.00\times9.8}{1.00+3.00}$

$a=2.45\ m/s^2$

We need to calculate the velocity

Using equation of motion

$v^2=u^2+2as$

Put the value into the formula

$v^2=0+2\times2.45\times1.00$

$v=\sqrt{2\times2.45\times1.00}$

$v=2.21\ m/s$

We need to calculate the kinetic energy of box b

Using formula of energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.00\times(2.21)^2$

$K.E=2.44\ J$

Hence, The kinetic energy of box b is 2.44 J.