An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 42KJ# over #t in [0,8s]#. What is the average speed of the object?

Question

An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 42KJ# over #t in [0,8s]#.  What is the average speed of the object?

Elena Albarran · Accepted Answer

The average speed is #=142.5 ms^(-1)# The kinetic energy #=(mv^2)/2# Let #u_0# be the initial velocity and #u_1# the final velocity. The average speed is #=(u_0+u_1)/2# #m*u_0^2/2=150*10^3# #:. u_0^2=150*10^3/5# #=>##u_0=sqrt(30*10^3)=173.2 ms^(-1)# #m*u_1^2/2=42*10^3# #:. u_1^2=42*10^3/5# #=>##u_1=sqrt(42*200)=91.7 ms^(-1)# The average speed is #(173.2+91.7)/2=142.5 ms^(-1)#

Axel Acevedo · Answer

undefined To find the average speed of the object, we need to use the formula for kinetic energy:

$KE = \frac{1}{2}mv^2$

Rearranging the formula to solve for velocity (v):

$v = \sqrt{\frac{2KE}{m}}$

We'll first find the velocity at the initial kinetic energy and then at the final kinetic energy. Then, we'll calculate the average speed by finding the average of these velocities.

For the initial kinetic energy:

$KE_1 = 150 \text{ kJ} = 150 \times 10^3 \text{ J}$

For the final kinetic energy:

$KE_2 = 42 \text{ kJ} = 42 \times 10^3 \text{ J}$

Using the formula for velocity:

$v_1 = \sqrt{\frac{2 \times 150 \times 10^3}{5}} \approx 77.46 \text{ m/s}$

$v_2 = \sqrt{\frac{2 \times 42 \times 10^3}{5}} \approx 49.29 \text{ m/s}$

Now, we'll find the average speed:

$v_{\text{average}} = \frac{v_1 + v_2}{2} = \frac{77.46 \text{ m/s} + 49.29 \text{ m/s}}{2} \approx 63.38 \text{ m/s}$