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Use [tex][tex]$KE=\frac{1}{2} mv^2$[/tex][/tex].

A boy weighing [tex]$\rightarrow$[/tex] kilograms is riding a skateboard. He's moving at 2 meters/second and has 40 joules of kinetic energy. If he doubles his speed, his kinetic energy at the faster speed is [tex]$\approx$[/tex] joules.

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Answer :

GinnyAnswer
Sure! Let's walk through a detailed, step-by-step solution for this problem.

1. Given Data:
- Kinetic Energy (KE) = 40 joules.
- Velocity (v) = 2 meters/second.

2. Determine the mass (m) of the boy:
We use the formula for kinetic energy:

[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]

Plugging in the values we know:

[tex]\[ 40 = \frac{1}{2} m (2)^2 \][/tex]

Simplify the expression:

[tex]\[ 40 = \frac{1}{2} m \cdot 4 \][/tex]

[tex]\[ 40 = 2m \][/tex]

Solve for [tex]\( m \)[/tex]:

[tex]\[ m = \frac{40}{2} = 20 \text{ kilograms} \][/tex]

Therefore, the boy weighs 20 kilograms.

3. Finding the kinetic energy at a faster speed:
Let's assume the faster speed to be 4 meters/second.

Again, we use the kinetic energy formula:

[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]

Now plug in the mass (m = 20 kilograms) and the new velocity (v = 4 meters/second):

[tex]\[ KE = \frac{1}{2} \cdot 20 \cdot (4)^2 \][/tex]

Simplify the expression:

[tex]\[ KE = \frac{1}{2} \cdot 20 \cdot 16 \][/tex]

[tex]\[ KE = \frac{1}{2} \cdot 320 \][/tex]

[tex]\[ KE = 160 \text{ joules} \][/tex]

Therefore, the boy's kinetic energy at the faster speed is 160 joules.

To sum up, the boy weighs 20 kilograms and his kinetic energy at the faster speed is approximately 160 joules.

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