A force of 30 N acts at an angle of 60° on a body of mass 6 kg initially at rest on a smooth horizontal plane. Find the distance covered in 10 seconds.

A force of 30 N acts at an angle of 60° on a body of mass 6 kg initially at rest on a smooth horizontal plane. Find the distance covered in 10 seconds.

Answer Details

To solve this problem, we need to use the equations of motion. Since the force is acting on the body, it will cause an acceleration in the direction of the force. The acceleration can be found using Newton's second law of motion: $$F = ma$$ where F is the force, m is the mass of the body, and a is the acceleration. In this case, the force is 30 N and the mass is 6 kg. Therefore, the acceleration can be found as: $$a = \frac{F}{m} = \frac{30}{6} = 5\text{ m/s}^2$$ Next, we can use the equations of motion to find the distance covered by the body in 10 seconds. Since the body starts from rest, we can use the equation: $$s = \frac{1}{2}at^2$$ where s is the distance covered, a is the acceleration, and t is the time. Substituting the values, we get: $$s = \frac{1}{2} \times 5 \times 10^2 = 125\text{ m}$$ However, the force is acting at an angle of 60 degrees to the horizontal. Therefore, only the horizontal component of the force will cause motion in the horizontal direction. The horizontal component of the force is: $$F_h = F \cos \theta = 30 \cos 60^\circ = 15\text{ N}$$ We can now find the acceleration in the horizontal direction as: $$a_h = \frac{F_h}{m} = \frac{15}{6} = 2.5\text{ m/s}^2$$ Using the equation of motion, the distance covered in 10 seconds in the horizontal direction is: $$s_h = \frac{1}{2} \times 2.5 \times 10^2 = 12.5\text{ m}$$ However, we also need to find the distance covered in the vertical direction. The vertical component of the force is: $$F_v = F \sin \theta = 30 \sin 60^\circ = 15\sqrt{3}\text{ N}$$ Since there is no force acting in the vertical direction, the body will move under the influence of gravity only. The acceleration due to gravity is 9.8 m/s^2. Therefore, the distance covered in the vertical direction in 10 seconds is: $$s_v = \frac{1}{2} \times 9.8 \times 10^2 = 490\text{ m}$$ The total distance covered by the body is the hypotenuse of the right-angled triangle formed by the horizontal and vertical distances: $$s_{total} = \sqrt{s_h^2 + s_v^2} = \sqrt{(12.5)^2 + (490)^2} \approx 490\text{ m}$$ Therefore, the answer is 125 m.