In the study of Physics, one of the fundamental concepts is Forces and Motion. Forces are pushes or pulls that can cause objects to move, change their speed, or change their direction. Understanding how forces affect motion is crucial for explaining the behavior of objects in the physical world.
A car traveling at a constant speed on a straight road is a practical example of Newton's First Law. When the car suddenly stops, the passengers tend to lurch forward due to their inertia.
$$F_{\text{net}} = 0$$
A force of 20N is applied to a box with a mass of 5kg. Calculate the acceleration of the box using Newton's Second Law.
$$F = ma$$ $$20 = 5a$$ $$a = \frac{20}{5} = 4 , \text{m/s}^2$$
When a person jumps off a boat onto the shore, the action force is the person pushing against the boat, and the reaction force is the boat pushing back on the person.
A block of mass 10kg is pushed along a rough surface with a force of 50N. If the coefficient of kinetic friction is 0.2, calculate the acceleration of the block.
$$F_{\text{friction}} = \mu_k \times F_{\text{normal}}$$ $$F_{\text{friction}} = 0.2 \times (10 \times 9.8) = 19.6 , \text{N}$$ $$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 50 - 19.6 = 30.4 , \text{N}$$ $$a = \frac{F_{\text{net}}}{m} = \frac{30.4}{10} = 3.04 , \text{m/s}^2$$
Calculate the weight of a 5kg object on Earth, where the acceleration due to gravity is $9.8 , \text{m/s}^2$.
$$\text{Weight} = m \times g = 5 \times 9.8 = 49 , \text{N}$$
A 2kg object is pushed with a force of 10N. If the coefficient of static friction is 0.3, determine if the object will move.
Answer: The maximum static frictional force is $F_{\text{friction}} = \mu_s \times F_{\text{normal}} = 0.3 \times (2 \times 9.8) = 5.88 , \text{N}$. Since the applied force is greater than the maximum static frictional force, the object will move.
Explain the difference between static and kinetic friction.
Answer: Static friction occurs when an object is at rest, preventing it from moving, while kinetic friction acts on moving objects.
A car accelerates from rest at 2 $\text{m/s}^2$ for 5 seconds. Calculate the final velocity of the car.
Answer: Using the equation $v = u + at$, where $u$ is the initial velocity (0 $\text{m/s}$), $a = 2 , \text{m/s}^2$, and $t = 5 , \text{s}$, $$v = 0 + 2 \times 5 = 10 , \text{m/s}$$
Two objects of masses 4kg and 6kg are placed 2m apart. Calculate the gravitational force between them if the gravitational constant is $6.67 \times 10^{-11} , \text{N m}^2/\text{kg}^2$.
Answer: The force of gravity is given by the equation $F = \frac{G \times m_1 \times m_2}{r^2}$: $$F = \frac{6.67 \times 10^{-11} \times 4 \times 6}{2^2} = 1.6 \times 10^{-10} , \text{N}$$
A block of mass 8kg is pulled along a horizontal surface with a force of 40N. If the coefficient of kinetic friction is 0.4, calculate the acceleration of the block.
Answer: The net force is $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 40 - (0.4 \times (8 \times 9.8)) = 40 - 31.36 = 8.64 , \text{N}$. Therefore, $a = \frac{F_{\text{net}}}{m} = \frac{8.64}{8} = 1.08 , \text{m/s}^2$
Create a free account to bookmark notes, download past papers, track your revision and get AI study help - free for Kenyan students.
Already have one? Log in
Join our list and we'll send fresh study notes and past papers straight to your inbox.