Q32E - University Physics with Modern Physics

Question

A 30.0-kg crate is initially moving with a velocity that has magnitude 3.90 m/s in a direction 37.0° west of north. How much work must be done on the crate to change its velocity to 5.62m/s in a direction 63.0° south of east?

Step-by-Step Solution

Answer

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

Step 1: Identification of given data

The given data can be listed below,

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

Step 2: Concept/Significance of kinetic energy

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.

Step 3: Determination of the work must be done on the crate to change its velocity to 5.62m/s in a direction 63.0° south of east

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,

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

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,

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

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