Question

select the equation that shows that 36 j of work is done when a 2.0 - kg block of ice is moved from rest to a speed.
$w=delta ke=\frac{1}{4}mv^{2}=\frac{1}{4}(2.0 kg)(6.0 m/s)^{2}=36 j$
$w=delta ke = mv^{2}=(2.0 kg)(6.0 m/s)^{2}=36 j$
$w=delta ke=\frac{1}{2}mv^{2}=\frac{1}{2}(2.0 kg)(6.0 m/s)^{2}=36 j$
$w=delta ke=\frac{1}{2}mv=\frac{1}{2}(2.0 kg)(6.0 m/s)=36 j$

Explanation:

Step1: Recall work - energy theorem

Work done $W$ on an object is equal to the change in its kinetic energy $\Delta KE$. The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass of the object and $v$ is its velocity.

Step2: Substitute given values

Given $m = 2.0\ kg$ and assume $v=6.0\ m/s$. Then $\Delta KE=\frac{1}{2}mv^{2}=\frac{1}{2}(2.0\ kg)(6.0\ m/s)^{2}=\frac{1}{2}\times2.0\times36\ J = 36\ J$.

Answer:

$W=\Delta KE=\frac{1}{2}mv^{2}=\frac{1}{2}(2.0\ kg)(6.0\ m/s)^{2}=36\ J$ (the third option)