Question

The table shows the mechanical energy and velocity of a rock that was thrown four times. The rock has a mass of 2.4 kg.

Answer 1

Answer:

Second trial

Explanation:

The table shows the mechanical energy and velocity of a rock that was thrown four times. The rock has a mass of 2.4 kg.

We have to find during which trial was the rock the highest above the ground.  

We know that mechanical energy is equal to the sum of kinetic energy and potential energy.  

• For first trial, kinetic energy is $E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times 2.4\times 7^2=58.8\ J$

So, PE = ME-KE

PE₁ = 32 - 58.8 = -26.8 J

• For second trial, kinetic energy is $E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times 2.4\times 2^2=4.8\ J$

So, PE = ME-KE

PE₂ = 34 - 4.8 = 29.2 J

• For third trial, kinetic energy is $E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times 2.4\times 6^2=43.2\ J$

So, PE = ME-KE

PE₃ = 35 - 43.2 = -8.2 J

• For fourth trial, kinetic energy is $E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times 2.4\times 3^2=10.8\ J$

So, PE = ME-KE

PE₄ = 31 - 10.8 = 20.2 J

We know that potential energy is directly proportional to the height from the surface. In second trial, the rock have maximum potential energy.

Hence, in second trial the rock is highest above the ground.

Answer 2

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