A 40.0 kg crate is initially moving with a velocity that has magnitude 4.09 m/s in a direction 37.0 ^\circ west of north. How much work must be done on the crate to change its velocity to 5.49 m/s in a direction 63.0 ^\circ south of east?

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To find the work done on the crate, we need to calculate the change in kinetic energy. The work done on an object is equal to the change in its kinetic energy.

First, let's find the initial and final velocities of the crate.

The initial velocity of the crate has a magnitude of 4.09 m/s and is directed 37.0° west of north. We can break this velocity into its components using trigonometry.

The vertical component of the initial velocity is given by v_y = v_initial * sin θ, where θ is the angle west of north.

v_y = 4.09 m/s * sin(37.0°)
= 2.46 m/s (northward)

The horizontal component of the initial velocity is given by v_x = v_initial * cos θ.

v_x = 4.09 m/s * cos(37.0°)
= 3.24 m/s (westward)

Similarly, let's calculate the final velocity components.

The vertical component of the final velocity is given by v_y = v_final * sin θ, where θ is the angle south of east.

v_y = 5.49 m/s * sin(63.0°)
= 4.86 m/s (southward)

The horizontal component of the final velocity is given by v_x = v_final * cos θ.

v_x = 5.49 m/s * cos(63.0°)
= 2.47 m/s (eastward)

Now, we can calculate the change in kinetic energy using the formula:

ΔKE = 0.5 * m * (v_final^2 - v_initial^2)

where m is the mass of the crate, v_initial is the initial velocity, and v_final is the final velocity.

ΔKE = 0.5 * 40.0 kg * [(2.47 m/s)^2 + (4.86 m/s)^2 - (3.24 m/s)^2 - (2.46 m/s)^2]

ΔKE = 0.5 * 40.0 kg * (6.09 m^2/s^2 + 23.62 m^2/s^2 - 10.50 m^2/s^2 - 6.05 m^2/s^2)

ΔKE = 0.5 * 40.0 kg * 13.16 m^2/s^2

ΔKE = 262.80 J

Therefore, the work done on the crate to change its velocity is 262.80 Joules.