What is the speed of a 0.148 kg baseball if its
kinetic energy is 107 J ?
Answer in units of m/s.

Question

Charles Aeschliman · Accepted Answer

#v ~~38.03# m/s

Given: #KE = 107# J; #m = 0.148# kg Formula for kinetic energy:#" "KE = 1/2 mv^2# The units of kinetic energy is Joules: #1 J = 1 N*m = 1 kg*m/s^2 * m = 1 (kg*m^2)/s^2# Substitute the knowns into the #KE# equation: #107 (kg*m^2)/s^2 = 1/2 (0.148 kg) v^2# Multiply both sides by #2#: #2/1 *107 (kg*m^2)/s^2 = 2/1 *1/2 (0.148 kg) v^2# #214 (kg*m^2)/s^2 = (0.148 kg) v^2# #(214 (cancel(kg)*m^2)/s^2)/ (0.148cancel( kg)) = v^2# #1445.95 m^2/s^2 = v^2# #v = sqrt(1445.95 m^2/s^2) ~~38.03 m/s#

Parker Abbott · Answer

undefined To find the speed of the baseball, we can use the formula for kinetic energy:

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

Where:
$ KE $ = Kinetic Energy (given as 107 J)
$ m $ = Mass of the baseball (given as 0.148 kg)
$ v $ = Speed of the baseball (what we need to find)

Rearranging the formula to solve for $ v $:

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

Substitute the given values:

$$ v = \sqrt{\frac{2 \times 107 \, \text{J}}{0.148 \, \text{kg}}} $$

$$ v = \sqrt{\frac{214 \, \text{J}}{0.148 \, \text{kg}}} $$

$$ v = \sqrt{1445.95} \, \text{m/s} $$

$$ v \approx 37.99 \, \text{m/s} $$

So, the speed of the baseball is approximately $ 37.99 \, \text{m/s} $.

What is the speed of a 0.148 kg baseball if its kinetic energy is 107 J ? Answer in units of m/s.