91影视

The given data can be listed below,

• The mass of the crate is,$m=30kg$ .
• The velocity of the crate is,$v_i=3.90m/s$ .
• The initial direction of the crate is,$胃=37掳$ .
• The final velocity of the crate is,$v_f=5.62m/s$ .

The input of a mass's free movement into a change in momentum, or the release of that mass's free movement into a change in momentum, is known as kinetic energy.

According to the total work theorem, which claims that the change in kinetic energy is equal to the entire work performed on or by the crate is given by,

$$\begin{gathered} W_{tot}=K_f-K_i \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \end{gathered}$$

Here, m is the mass of the crate and$v_f$ is the final velocity of the crate,$v_1$ is the initial velocity of the crate.

Substitute all the values in the above,

$$\begin{gathered} W_{tot}=\frac{1}{2}m\left(v_f^2-v_i^2\right) \\ =\frac{1}{2}\left(30kg\right)\left({\left(5.62\right)}^2-{\left(3.90\right)}^2\right)\left(m/s^2\right) \\ =\frac{1}{2}\left(30kg\right)\left(31.58-15.21\right){\left(m/s\right)}^2 \\ =245.6J \end{gathered}$$

Thus, the work done on the crate to change its velocity is 245.6 J .