Answer :
Let's tackle each part of your question step by step.
Which of the following is/are correct about Newton's laws of motion?
To determine the correct statements, let's review Newton's laws of motion:
I. Newton's first law of motion is otherwise known as the law of inertia.
- This is true. Newton's first law states that an object at rest stays at rest, and an object in motion continues in motion with the same speed and in the same direction unless acted upon by a net external force. This property is known as inertia.
II. The rate of change of linear momentum is not directly proportional to the applied force.
- This is false. According to Newton's second law, the rate of change of momentum of an object is directly proportional to the net applied force.
III. Ft = impulse.
- This is true. Impulse is defined as the product of the force applied and the time duration for which it acts. It is mathematically represented as [tex]F \times t[/tex].
Therefore, the correct answer is B. I and III only.
An object of mass 10 kg rests on a frictionless horizontal surface. How large an acceleration will it undergo if a horizontal force of 12 N acts on it?
To find the acceleration, use Newton's second law of motion, [tex]F = ma[/tex].
Given that:
- Mass [tex]m = 10 \text{ kg}[/tex]
- Force [tex]F = 12 \text{ N}[/tex]
Rearrange the formula to find [tex]a[/tex]:
[tex]a = \frac{F}{m} = \frac{12 \text{ N}}{10 \text{ kg}} = 1.2 \text{ m/s}^2[/tex]The correct answer is B. 1.2 m/s².
A body of 6 kg undergoes a constant horizontal acceleration of 2 m/s². Calculate the resultant horizontal force acting on the body. What will be the resultant force on the body when it moves with a uniform velocity of 5 m/s?
For the first part, use Newton's second law:
- Mass [tex]m = 6 \text{ kg}[/tex]
- Acceleration [tex]a = 2 \text{ m/s}^2[/tex]
[tex]F = ma = 6 \text{ kg} \times 2 \text{ m/s}^2 = 12 \text{ N}[/tex]
When the body moves with a uniform velocity, it means there's no net external force acting on it, so:
- Resultant force [tex]= 0 \text{ N}[/tex] (since acceleration is 0)
The correct answer is A. 12 N, 0 N.
A 10 kg block is at rest. A constant force F acts on the block along the x-axis. At time t = 2 seconds, the block moves 8 m. Find F?
We can use the equation of motion [tex]s = ut + \frac{1}{2}at^2[/tex]. Since the block is initially at rest, initial velocity [tex]u = 0[/tex].
Given:
- Displacement [tex]s = 8 \text{ m}[/tex]
- Time [tex]t = 2 \text{ s}[/tex]
Plug these values into the equation:
[tex]8 = 0 \times 2 + \frac{1}{2}a (2)^2[/tex]
[tex]8 = 2a[/tex]
[tex]a = 4 \text{ m/s}^2[/tex]Now, use Newton’s second law to find [tex]F[/tex]:
- Mass [tex]m = 10 \text{ kg}[/tex]
[tex]F = ma = 10 \text{ kg} \times 4 \text{ m/s}^2 = 40 \text{ N}[/tex]
The correct answer is B. 40 N.
