Question: A 2 kg block is initially at rest. A force of 10 N is applied to the block for a distance of 3 meters along a frictionless surface, at an angle of 30 degrees above the horizontal. Calculate the work done by the force and the block's final velocity after the force is applied.

Answer:

1. Work Done by the Force: The work done by the force can be calculated using the formula: [W = Fd\cos\theta] Where, $W$ = work done by the force (in joules) $F$ = magnitude of the force (in newtons) $d$ = distance over which the force is applied (in meters) $\theta$ = angle between the force and the direction of the displacement Plugging in the given values: [W = 10 \times 3 \times \cos(30^\circ)] [W = 10 \times 3 \times \frac{\sqrt{3}}{2}] [W = 15\sqrt{3}] [W \approx 25.98 J]

2. Final Velocity of the Block: The work done by the external force results in a change in the kinetic energy of the block: [W = \Delta KE] Using the work-energy principle and the fact that the initial kinetic energy is zero, we can find the final kinetic energy: [W = KE_f - KE_i] [KE_f = W + KE_i] [KE_f = 25.98 + 0] [KE_f = 25.98 J] Now, the kinetic energy at the final velocity can be expressed as: [KE_f = \frac{1}{2}mv^2] Where, $m$ = mass of the block (in kilograms) $v$ = final velocity of the block (in meters per second) Rearranging the formula: [v = \sqrt{\frac{2KE_f}{m}}] [v = \sqrt{\frac{2 \times 25.98}{2}}] [v = \sqrt{25.98}] [v \approx 5.10 m/s]

Therefore, the work done by the force is approximately 25.98 J, and the block's final velocity after the force is applied is approximately 5.10 m/s.